Show that this matrix is not diagonalizable

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Say I have a matrix:



$$A = beginbmatrix 2 & 0 \ -1 & 2 endbmatrix $$



  1. Is this matrix diagonalizable?


  2. Does a 2x2 matrix always have 2 eigenvalues (multipicity counts). Why is this? I know this matrix (because it's lower triangular) has the eigenvalue of 2 with multiplicity 2... but does a matrix of this size always have 2 eigenvalues. Why is this?


Is there any way to know if the eigenvalue of 2 has two eigenvectors or not quickly? Here's the way I know to find the eigenvector:



$$beginbmatrix 2 & 0 \ 0 & 2 endbmatrix - beginbmatrix 2 & 0 \ -1 & 2 endbmatrix = beginbmatrix 0 & 0 \ 1 & 0 endbmatrix $$



$$ eigenvector = beginbmatrix x_1 \ x_2 endbmatrix = beginbmatrix 0 \ t endbmatrix = t * beginbmatrix 0 \ 1 endbmatrix$$



By this theorem, it is not diagonalizable because it only has 1 eigenvector right and the matrix has 2 rows and 2 columns:



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linear-algebra eigenvalues-eigenvectors diagonalization







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  • You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
    – hardmath
    2 hours ago














up vote
1
down vote

favorite












Say I have a matrix:



$$A = beginbmatrix 2 & 0 \ -1 & 2 endbmatrix $$



  1. Is this matrix diagonalizable?


  2. Does a 2x2 matrix always have 2 eigenvalues (multipicity counts). Why is this? I know this matrix (because it's lower triangular) has the eigenvalue of 2 with multiplicity 2... but does a matrix of this size always have 2 eigenvalues. Why is this?


Is there any way to know if the eigenvalue of 2 has two eigenvectors or not quickly? Here's the way I know to find the eigenvector:



$$beginbmatrix 2 & 0 \ 0 & 2 endbmatrix - beginbmatrix 2 & 0 \ -1 & 2 endbmatrix = beginbmatrix 0 & 0 \ 1 & 0 endbmatrix $$



$$ eigenvector = beginbmatrix x_1 \ x_2 endbmatrix = beginbmatrix 0 \ t endbmatrix = t * beginbmatrix 0 \ 1 endbmatrix$$



By this theorem, it is not diagonalizable because it only has 1 eigenvector right and the matrix has 2 rows and 2 columns:



enter image description here






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question









New contributor




Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
    – hardmath
    2 hours ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Say I have a matrix:



$$A = beginbmatrix 2 & 0 \ -1 & 2 endbmatrix $$



  1. Is this matrix diagonalizable?


  2. Does a 2x2 matrix always have 2 eigenvalues (multipicity counts). Why is this? I know this matrix (because it's lower triangular) has the eigenvalue of 2 with multiplicity 2... but does a matrix of this size always have 2 eigenvalues. Why is this?


Is there any way to know if the eigenvalue of 2 has two eigenvectors or not quickly? Here's the way I know to find the eigenvector:



$$beginbmatrix 2 & 0 \ 0 & 2 endbmatrix - beginbmatrix 2 & 0 \ -1 & 2 endbmatrix = beginbmatrix 0 & 0 \ 1 & 0 endbmatrix $$



$$ eigenvector = beginbmatrix x_1 \ x_2 endbmatrix = beginbmatrix 0 \ t endbmatrix = t * beginbmatrix 0 \ 1 endbmatrix$$



By this theorem, it is not diagonalizable because it only has 1 eigenvector right and the matrix has 2 rows and 2 columns:



enter image description here






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question









New contributor




Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Say I have a matrix:



$$A = beginbmatrix 2 & 0 \ -1 & 2 endbmatrix $$



  1. Is this matrix diagonalizable?


  2. Does a 2x2 matrix always have 2 eigenvalues (multipicity counts). Why is this? I know this matrix (because it's lower triangular) has the eigenvalue of 2 with multiplicity 2... but does a matrix of this size always have 2 eigenvalues. Why is this?


Is there any way to know if the eigenvalue of 2 has two eigenvectors or not quickly? Here's the way I know to find the eigenvector:



$$beginbmatrix 2 & 0 \ 0 & 2 endbmatrix - beginbmatrix 2 & 0 \ -1 & 2 endbmatrix = beginbmatrix 0 & 0 \ 1 & 0 endbmatrix $$



$$ eigenvector = beginbmatrix x_1 \ x_2 endbmatrix = beginbmatrix 0 \ t endbmatrix = t * beginbmatrix 0 \ 1 endbmatrix$$



By this theorem, it is not diagonalizable because it only has 1 eigenvector right and the matrix has 2 rows and 2 columns:



enter image description here






linear-algebra eigenvalues-eigenvectors diagonalization




linear-algebra eigenvalues-eigenvectors diagonalization






share|cite|improve this question









New contributor




Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited 2 hours ago









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Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
    – hardmath
    2 hours ago
















  • You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
    – hardmath
    2 hours ago















You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
– hardmath
2 hours ago




You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
– hardmath
2 hours ago










3 Answers
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  1. No, it's not diagonalizable. If the two eigenvalues of a $ 2 times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).


  2. The eigenvalues of an $n times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n times n$ matrix has $n$ eigenvalues (when counted with multiplicities).


By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.






share|cite|improve this answer



























    up vote
    1
    down vote













    HINT



    Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.






    share|cite|improve this answer



























      up vote
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      down vote













      1. No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)

      2. To give a quick example, let's consider the 2D rotation matrix. $$beginpmatrixcos theta &-sin theta \sin theta &cos theta \endpmatrix$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.





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        3 Answers
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        1. No, it's not diagonalizable. If the two eigenvalues of a $ 2 times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).


        2. The eigenvalues of an $n times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n times n$ matrix has $n$ eigenvalues (when counted with multiplicities).


        By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.






        share|cite|improve this answer
























          up vote
          4
          down vote



          accepted










          1. No, it's not diagonalizable. If the two eigenvalues of a $ 2 times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).


          2. The eigenvalues of an $n times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n times n$ matrix has $n$ eigenvalues (when counted with multiplicities).


          By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.






          share|cite|improve this answer






















            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            1. No, it's not diagonalizable. If the two eigenvalues of a $ 2 times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).


            2. The eigenvalues of an $n times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n times n$ matrix has $n$ eigenvalues (when counted with multiplicities).


            By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.






            share|cite|improve this answer












            1. No, it's not diagonalizable. If the two eigenvalues of a $ 2 times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).


            2. The eigenvalues of an $n times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n times n$ matrix has $n$ eigenvalues (when counted with multiplicities).


            By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            John Hughes

            60.1k23987




            60.1k23987




















                up vote
                1
                down vote













                HINT



                Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.






                share|cite|improve this answer
























                  up vote
                  1
                  down vote













                  HINT



                  Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    HINT



                    Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.






                    share|cite|improve this answer












                    HINT



                    Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    gimusi

                    75.9k73889




                    75.9k73889




















                        up vote
                        1
                        down vote













                        1. No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)

                        2. To give a quick example, let's consider the 2D rotation matrix. $$beginpmatrixcos theta &-sin theta \sin theta &cos theta \endpmatrix$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.





                        share|cite|improve this answer
























                          up vote
                          1
                          down vote













                          1. No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)

                          2. To give a quick example, let's consider the 2D rotation matrix. $$beginpmatrixcos theta &-sin theta \sin theta &cos theta \endpmatrix$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.





                          share|cite|improve this answer






















                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            1. No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)

                            2. To give a quick example, let's consider the 2D rotation matrix. $$beginpmatrixcos theta &-sin theta \sin theta &cos theta \endpmatrix$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.





                            share|cite|improve this answer












                            1. No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)

                            2. To give a quick example, let's consider the 2D rotation matrix. $$beginpmatrixcos theta &-sin theta \sin theta &cos theta \endpmatrix$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.






                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









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