Mixing
A question of non-singularity
Clash Royale CLAN TAG#URR8PPP
up vote
2
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Let $A$ and $B$ be matrices such that $B^2+ AB + 2I = 0$, where I
denotes the identity matrix. Which of the following matrices must be
nonsingular?
(A) $A + 2I$
(B) $B$
(C) $B + 2I$
(D) $A$
I tried using a few tricks assuming each option to be nonsingular and then coming to the given form but to no avail. Any hint is appreciated.
linear-algebra matrices determinant
edited 3 hours ago
greedoid
30.6k94184
asked 3 hours ago
Kc17
245
add a comment |Â
up vote
2
down vote
favorite
Let $A$ and $B$ be matrices such that $B^2+ AB + 2I = 0$, where I
denotes the identity matrix. Which of the following matrices must be
nonsingular?
(A) $A + 2I$
(B) $B$
(C) $B + 2I$
(D) $A$
I tried using a few tricks assuming each option to be nonsingular and then coming to the given form but to no avail. Any hint is appreciated.
linear-algebra matrices determinant
edited 3 hours ago
greedoid
30.6k94184
asked 3 hours ago
Kc17
245
I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
– Robert Lewis
3 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $A$ and $B$ be matrices such that $B^2+ AB + 2I = 0$, where I
denotes the identity matrix. Which of the following matrices must be
nonsingular?
(A) $A + 2I$
(B) $B$
(C) $B + 2I$
(D) $A$
I tried using a few tricks assuming each option to be nonsingular and then coming to the given form but to no avail. Any hint is appreciated.
linear-algebra matrices determinant
edited 3 hours ago
greedoid
30.6k94184
asked 3 hours ago
Kc17
245
Let $A$ and $B$ be matrices such that $B^2+ AB + 2I = 0$, where I
denotes the identity matrix. Which of the following matrices must be
nonsingular?
(A) $A + 2I$
(B) $B$
(C) $B + 2I$
(D) $A$
I tried using a few tricks assuming each option to be nonsingular and then coming to the given form but to no avail. Any hint is appreciated.
linear-algebra matrices determinant
linear-algebra matrices determinant
edited 3 hours ago
greedoid
30.6k94184
asked 3 hours ago
Kc17
245
edited 3 hours ago
greedoid
30.6k94184
asked 3 hours ago
Kc17
245
edited 3 hours ago
greedoid
30.6k94184
edited 3 hours ago
greedoid
30.6k94184
edited 3 hours ago
greedoid
30.6k94184
30.6k94184
asked 3 hours ago
Kc17
245
asked 3 hours ago
Kc17
245
asked 3 hours ago
Kc17
245
245
I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
– Robert Lewis
3 hours ago
add a comment |Â
I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
– Robert Lewis
3 hours ago
I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
– Robert Lewis
3 hours ago
I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
– Robert Lewis
3 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
Since $$det (B+A)cdot det (B) = det (B^2+AB) = det (-2I) =-2$$ we see that $det (B)ne 0$ so $B$ is invertibile.
answered 3 hours ago
greedoid
30.6k94184
add a comment |Â
up vote
2
down vote
We have
$B^2 + AB + 2I = 0, tag 1$
so
$(B + A)B = B^2 + AB = -2I, tag 2$
or
$left ( -dfrac12(B + A) right )B = I, tag 3$
that is,
$B^-1 = -dfrac12(B + A), tag 4$
and the correct answer is (B).
answered 3 hours ago
Robert Lewis
39.7k22459
@StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
– Robert Lewis
2 hours ago
1
@StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
– Robert Lewis
2 hours ago
How do you exclude the other cases?
– egreg
50 mins ago
add a comment |Â
up vote
0
down vote
I'm assuming you're over the real numbers.
Suppose $B$ is singular and take $vne0$ such that $Bv=0$; then
$$
0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
$$
a contradiction. Therefore $B$ is nonsingular.
In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.
The matrix $A$ can be the zero matrix, because
$$
beginbmatrix
0 & 2 \
-1 & 0
endbmatrix^2=-2I
$$
Also $A+2I$ can be the zero matrix: the identity to satisfy would be
$$
B^2-2B+2I=0
$$
and the matrix
beginbmatrix
2 & -1 \
2 & 0
endbmatrix
satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.
Can $B+2I=0$?
answered 51 mins ago
egreg
169k1283191
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Since $$det (B+A)cdot det (B) = det (B^2+AB) = det (-2I) =-2$$ we see that $det (B)ne 0$ so $B$ is invertibile.
answered 3 hours ago
greedoid
30.6k94184
add a comment |Â
up vote
4
down vote
Since $$det (B+A)cdot det (B) = det (B^2+AB) = det (-2I) =-2$$ we see that $det (B)ne 0$ so $B$ is invertibile.
answered 3 hours ago
greedoid
30.6k94184
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Since $$det (B+A)cdot det (B) = det (B^2+AB) = det (-2I) =-2$$ we see that $det (B)ne 0$ so $B$ is invertibile.
answered 3 hours ago
greedoid
30.6k94184
Since $$det (B+A)cdot det (B) = det (B^2+AB) = det (-2I) =-2$$ we see that $det (B)ne 0$ so $B$ is invertibile.
answered 3 hours ago
greedoid
30.6k94184
answered 3 hours ago
greedoid
30.6k94184
answered 3 hours ago
greedoid
30.6k94184
answered 3 hours ago
greedoid
30.6k94184
30.6k94184
add a comment |Â
add a comment |Â
up vote
2
down vote
We have
$B^2 + AB + 2I = 0, tag 1$
so
$(B + A)B = B^2 + AB = -2I, tag 2$
or
$left ( -dfrac12(B + A) right )B = I, tag 3$
that is,
$B^-1 = -dfrac12(B + A), tag 4$
and the correct answer is (B).
answered 3 hours ago
Robert Lewis
39.7k22459
@StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
– Robert Lewis
2 hours ago
1
@StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
– Robert Lewis
2 hours ago
How do you exclude the other cases?
– egreg
50 mins ago
add a comment |Â
up vote
2
down vote
We have
$B^2 + AB + 2I = 0, tag 1$
so
$(B + A)B = B^2 + AB = -2I, tag 2$
or
$left ( -dfrac12(B + A) right )B = I, tag 3$
that is,
$B^-1 = -dfrac12(B + A), tag 4$
and the correct answer is (B).
answered 3 hours ago
Robert Lewis
39.7k22459
@StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
– Robert Lewis
2 hours ago
1
@StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
– Robert Lewis
2 hours ago
How do you exclude the other cases?
– egreg
50 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have
$B^2 + AB + 2I = 0, tag 1$
so
$(B + A)B = B^2 + AB = -2I, tag 2$
or
$left ( -dfrac12(B + A) right )B = I, tag 3$
that is,
$B^-1 = -dfrac12(B + A), tag 4$
and the correct answer is (B).
answered 3 hours ago
Robert Lewis
39.7k22459
We have
$B^2 + AB + 2I = 0, tag 1$
so
$(B + A)B = B^2 + AB = -2I, tag 2$
or
$left ( -dfrac12(B + A) right )B = I, tag 3$
that is,
$B^-1 = -dfrac12(B + A), tag 4$
and the correct answer is (B).
answered 3 hours ago
Robert Lewis
39.7k22459
edited 2 hours ago
answered 3 hours ago
Robert Lewis
39.7k22459
answered 3 hours ago
Robert Lewis
39.7k22459
answered 3 hours ago
Robert Lewis
39.7k22459
39.7k22459
@StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
– Robert Lewis
2 hours ago
1
@StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
– Robert Lewis
2 hours ago
How do you exclude the other cases?
– egreg
50 mins ago
add a comment |Â
@StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
– Robert Lewis
2 hours ago
1
@StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
– Robert Lewis
2 hours ago
How do you exclude the other cases?
– egreg
50 mins ago
@StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
– Robert Lewis
2 hours ago
@StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
– Robert Lewis
2 hours ago
1
1
@StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
– Robert Lewis
2 hours ago
@StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
– Robert Lewis
2 hours ago
How do you exclude the other cases?
– egreg
50 mins ago
How do you exclude the other cases?
– egreg
50 mins ago
add a comment |Â
up vote
0
down vote
I'm assuming you're over the real numbers.
Suppose $B$ is singular and take $vne0$ such that $Bv=0$; then
$$
0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
$$
a contradiction. Therefore $B$ is nonsingular.
In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.
The matrix $A$ can be the zero matrix, because
$$
beginbmatrix
0 & 2 \
-1 & 0
endbmatrix^2=-2I
$$
Also $A+2I$ can be the zero matrix: the identity to satisfy would be
$$
B^2-2B+2I=0
$$
and the matrix
beginbmatrix
2 & -1 \
2 & 0
endbmatrix
satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.
Can $B+2I=0$?
answered 51 mins ago
egreg
169k1283191
add a comment |Â
up vote
0
down vote
I'm assuming you're over the real numbers.
Suppose $B$ is singular and take $vne0$ such that $Bv=0$; then
$$
0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
$$
a contradiction. Therefore $B$ is nonsingular.
In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.
The matrix $A$ can be the zero matrix, because
$$
beginbmatrix
0 & 2 \
-1 & 0
endbmatrix^2=-2I
$$
Also $A+2I$ can be the zero matrix: the identity to satisfy would be
$$
B^2-2B+2I=0
$$
and the matrix
beginbmatrix
2 & -1 \
2 & 0
endbmatrix
satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.
Can $B+2I=0$?
answered 51 mins ago
egreg
169k1283191
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I'm assuming you're over the real numbers.
Suppose $B$ is singular and take $vne0$ such that $Bv=0$; then
$$
0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
$$
a contradiction. Therefore $B$ is nonsingular.
In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.
The matrix $A$ can be the zero matrix, because
$$
beginbmatrix
0 & 2 \
-1 & 0
endbmatrix^2=-2I
$$
Also $A+2I$ can be the zero matrix: the identity to satisfy would be
$$
B^2-2B+2I=0
$$
and the matrix
beginbmatrix
2 & -1 \
2 & 0
endbmatrix
satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.
Can $B+2I=0$?
answered 51 mins ago
egreg
169k1283191
I'm assuming you're over the real numbers.
Suppose $B$ is singular and take $vne0$ such that $Bv=0$; then
$$
0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
$$
a contradiction. Therefore $B$ is nonsingular.
In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.
The matrix $A$ can be the zero matrix, because
$$
beginbmatrix
0 & 2 \
-1 & 0
endbmatrix^2=-2I
$$
Also $A+2I$ can be the zero matrix: the identity to satisfy would be
$$
B^2-2B+2I=0
$$
and the matrix
beginbmatrix
2 & -1 \
2 & 0
endbmatrix
satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.
Can $B+2I=0$?
answered 51 mins ago
egreg
169k1283191
answered 51 mins ago
egreg
169k1283191
answered 51 mins ago
egreg
169k1283191
answered 51 mins ago
egreg
169k1283191
169k1283191
add a comment |Â
add a comment |Â
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I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
– Robert Lewis
3 hours ago