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Operator Norm from Geometrical Point of View
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I'm trying to understand some Functional Analysis concepts from Geometrical point of view. So I know that Metric ($d(a,b)$) is the distance between elements, Norm ($||a||$) is the length of element.
So I'd like to know how Operator Norm ($||A||$, $A: X to Y$) could be represented as Geometrical concept.
geometry functional-analysis normed-spaces
asked Aug 23 at 7:14
Denis Sablukov
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up vote
2
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I'm trying to understand some Functional Analysis concepts from Geometrical point of view. So I know that Metric ($d(a,b)$) is the distance between elements, Norm ($||a||$) is the length of element.
So I'd like to know how Operator Norm ($||A||$, $A: X to Y$) could be represented as Geometrical concept.
geometry functional-analysis normed-spaces
asked Aug 23 at 7:14
Denis Sablukov
165
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to understand some Functional Analysis concepts from Geometrical point of view. So I know that Metric ($d(a,b)$) is the distance between elements, Norm ($||a||$) is the length of element.
So I'd like to know how Operator Norm ($||A||$, $A: X to Y$) could be represented as Geometrical concept.
geometry functional-analysis normed-spaces
asked Aug 23 at 7:14
Denis Sablukov
165
I'm trying to understand some Functional Analysis concepts from Geometrical point of view. So I know that Metric ($d(a,b)$) is the distance between elements, Norm ($||a||$) is the length of element.
So I'd like to know how Operator Norm ($||A||$, $A: X to Y$) could be represented as Geometrical concept.
geometry functional-analysis normed-spaces
asked Aug 23 at 7:14
Denis Sablukov
165
asked Aug 23 at 7:14
Denis Sablukov
165
asked Aug 23 at 7:14
Denis Sablukov
165
asked Aug 23 at 7:14
Denis Sablukov
165
165
add a comment |Â
add a comment |Â
2 Answers
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If $A:Xto Y$ is a linear bounded operator between the normed spaces $X,Y$, then the norm of $A$ is, loosely speaking, the factor by which the unit ball of $X$ gets inflated or deflated. More precisely, it is the smallest number $alpha>0$ by which you need to dilate the unit ball of $Y$, so that $alpha B_Y$ will contain the image of the unit ball of $X$ under the mapping $A$.
For example, a diagonal matrix $D=d_i_i=1^n$ in $mathbbR^n$ maps the Euclidean unit ball to an ellipsoid, whose principal axis have lengths $2|d_i|$. The maximum of $|d_i|$ is of course the norm of this diagonal operator as a map between $ell_2^n$ and itself. It is also the smallest number by which you need to multiply the unit ball of $ell_2^n$ in order that the inflated ball will contain the ellipsoid $D(B_2^n)$, where $B_2^n$ is the unit ball of $ell_2^n$.
answered Aug 23 at 7:25
uniquesolution
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Take the (closed) unit ball $B$. It is all the points of norm at most $1$.
Now, apply the operator $T$ to the unit ball, to get a convex set $T(B)$. The operator norm of $T$ is the radius of the smallest ball centered at $0$ that will contain $T(B)$. In finite dimensions, you think of $T$ turning the unit disk into some sort of ellipse-like region and the maximal distance from the origin of the boundary of this region is the operator norm.
For example, if $|T| = 0$, then that means that $T$ sends $B$ to the single point $0$, because the smallest closed ball containing $T(B)$ is in fact just the set $0$. We then have by linearity that $T$ is identically zero.
This is why the inequality $|TS| leq |T||S|$ is actually intuitive. Let $r$ be the radius of the smallest closed ball containing $TS(B)$ (i.e. the operator norm of $TS$). If we look at the smallest radius that contains $S(B)$, $s$, and apply $T$ to that radius ball and ask what is the smallest ball containing that set, get $|T||S|$, but since this set clearly contains $TS(B)$, because instead of $T(S(B))$, we took $T(sB)$ and $S(B) subset sB$, by definition of $s$.
Likewise, you can check geometrically the triangle inequality: $|T + S| leq |T| + |S|$, because $(T + S)(B) subset T(B) + S(B)$, because the first depends on adding only the points of the image that come from the same vector (i.e. $T(v) + S(v)$, but not $T(v) + S(w)$ if $v neq w$) and the latter adds using all vectors in $B$ for arguments of $T$ and any vector in $B$ for an argument for $S$. Then we see that if $t$ is the smallest radius containing $T(B)$ and $s$ is the smallest radius contain $S(B)$, then the smallest radius containing $T(B) + S(B)$ is at most $t+s$, because $tB + sB = (t+s)B$, as sets.
answered Aug 23 at 7:32
4-ier
6089
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
If $A:Xto Y$ is a linear bounded operator between the normed spaces $X,Y$, then the norm of $A$ is, loosely speaking, the factor by which the unit ball of $X$ gets inflated or deflated. More precisely, it is the smallest number $alpha>0$ by which you need to dilate the unit ball of $Y$, so that $alpha B_Y$ will contain the image of the unit ball of $X$ under the mapping $A$.
For example, a diagonal matrix $D=d_i_i=1^n$ in $mathbbR^n$ maps the Euclidean unit ball to an ellipsoid, whose principal axis have lengths $2|d_i|$. The maximum of $|d_i|$ is of course the norm of this diagonal operator as a map between $ell_2^n$ and itself. It is also the smallest number by which you need to multiply the unit ball of $ell_2^n$ in order that the inflated ball will contain the ellipsoid $D(B_2^n)$, where $B_2^n$ is the unit ball of $ell_2^n$.
answered Aug 23 at 7:25
uniquesolution
8,271823
add a comment |Â
up vote
8
down vote
accepted
If $A:Xto Y$ is a linear bounded operator between the normed spaces $X,Y$, then the norm of $A$ is, loosely speaking, the factor by which the unit ball of $X$ gets inflated or deflated. More precisely, it is the smallest number $alpha>0$ by which you need to dilate the unit ball of $Y$, so that $alpha B_Y$ will contain the image of the unit ball of $X$ under the mapping $A$.
For example, a diagonal matrix $D=d_i_i=1^n$ in $mathbbR^n$ maps the Euclidean unit ball to an ellipsoid, whose principal axis have lengths $2|d_i|$. The maximum of $|d_i|$ is of course the norm of this diagonal operator as a map between $ell_2^n$ and itself. It is also the smallest number by which you need to multiply the unit ball of $ell_2^n$ in order that the inflated ball will contain the ellipsoid $D(B_2^n)$, where $B_2^n$ is the unit ball of $ell_2^n$.
answered Aug 23 at 7:25
uniquesolution
8,271823
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
If $A:Xto Y$ is a linear bounded operator between the normed spaces $X,Y$, then the norm of $A$ is, loosely speaking, the factor by which the unit ball of $X$ gets inflated or deflated. More precisely, it is the smallest number $alpha>0$ by which you need to dilate the unit ball of $Y$, so that $alpha B_Y$ will contain the image of the unit ball of $X$ under the mapping $A$.
For example, a diagonal matrix $D=d_i_i=1^n$ in $mathbbR^n$ maps the Euclidean unit ball to an ellipsoid, whose principal axis have lengths $2|d_i|$. The maximum of $|d_i|$ is of course the norm of this diagonal operator as a map between $ell_2^n$ and itself. It is also the smallest number by which you need to multiply the unit ball of $ell_2^n$ in order that the inflated ball will contain the ellipsoid $D(B_2^n)$, where $B_2^n$ is the unit ball of $ell_2^n$.
answered Aug 23 at 7:25
uniquesolution
8,271823
If $A:Xto Y$ is a linear bounded operator between the normed spaces $X,Y$, then the norm of $A$ is, loosely speaking, the factor by which the unit ball of $X$ gets inflated or deflated. More precisely, it is the smallest number $alpha>0$ by which you need to dilate the unit ball of $Y$, so that $alpha B_Y$ will contain the image of the unit ball of $X$ under the mapping $A$.
For example, a diagonal matrix $D=d_i_i=1^n$ in $mathbbR^n$ maps the Euclidean unit ball to an ellipsoid, whose principal axis have lengths $2|d_i|$. The maximum of $|d_i|$ is of course the norm of this diagonal operator as a map between $ell_2^n$ and itself. It is also the smallest number by which you need to multiply the unit ball of $ell_2^n$ in order that the inflated ball will contain the ellipsoid $D(B_2^n)$, where $B_2^n$ is the unit ball of $ell_2^n$.
answered Aug 23 at 7:25
uniquesolution
8,271823
answered Aug 23 at 7:25
uniquesolution
8,271823
answered Aug 23 at 7:25
uniquesolution
8,271823
answered Aug 23 at 7:25
uniquesolution
8,271823
8,271823
add a comment |Â
add a comment |Â
up vote
5
down vote
Take the (closed) unit ball $B$. It is all the points of norm at most $1$.
Now, apply the operator $T$ to the unit ball, to get a convex set $T(B)$. The operator norm of $T$ is the radius of the smallest ball centered at $0$ that will contain $T(B)$. In finite dimensions, you think of $T$ turning the unit disk into some sort of ellipse-like region and the maximal distance from the origin of the boundary of this region is the operator norm.
For example, if $|T| = 0$, then that means that $T$ sends $B$ to the single point $0$, because the smallest closed ball containing $T(B)$ is in fact just the set $0$. We then have by linearity that $T$ is identically zero.
This is why the inequality $|TS| leq |T||S|$ is actually intuitive. Let $r$ be the radius of the smallest closed ball containing $TS(B)$ (i.e. the operator norm of $TS$). If we look at the smallest radius that contains $S(B)$, $s$, and apply $T$ to that radius ball and ask what is the smallest ball containing that set, get $|T||S|$, but since this set clearly contains $TS(B)$, because instead of $T(S(B))$, we took $T(sB)$ and $S(B) subset sB$, by definition of $s$.
Likewise, you can check geometrically the triangle inequality: $|T + S| leq |T| + |S|$, because $(T + S)(B) subset T(B) + S(B)$, because the first depends on adding only the points of the image that come from the same vector (i.e. $T(v) + S(v)$, but not $T(v) + S(w)$ if $v neq w$) and the latter adds using all vectors in $B$ for arguments of $T$ and any vector in $B$ for an argument for $S$. Then we see that if $t$ is the smallest radius containing $T(B)$ and $s$ is the smallest radius contain $S(B)$, then the smallest radius containing $T(B) + S(B)$ is at most $t+s$, because $tB + sB = (t+s)B$, as sets.
answered Aug 23 at 7:32
4-ier
6089
add a comment |Â
up vote
5
down vote
Take the (closed) unit ball $B$. It is all the points of norm at most $1$.
Now, apply the operator $T$ to the unit ball, to get a convex set $T(B)$. The operator norm of $T$ is the radius of the smallest ball centered at $0$ that will contain $T(B)$. In finite dimensions, you think of $T$ turning the unit disk into some sort of ellipse-like region and the maximal distance from the origin of the boundary of this region is the operator norm.
For example, if $|T| = 0$, then that means that $T$ sends $B$ to the single point $0$, because the smallest closed ball containing $T(B)$ is in fact just the set $0$. We then have by linearity that $T$ is identically zero.
This is why the inequality $|TS| leq |T||S|$ is actually intuitive. Let $r$ be the radius of the smallest closed ball containing $TS(B)$ (i.e. the operator norm of $TS$). If we look at the smallest radius that contains $S(B)$, $s$, and apply $T$ to that radius ball and ask what is the smallest ball containing that set, get $|T||S|$, but since this set clearly contains $TS(B)$, because instead of $T(S(B))$, we took $T(sB)$ and $S(B) subset sB$, by definition of $s$.
Likewise, you can check geometrically the triangle inequality: $|T + S| leq |T| + |S|$, because $(T + S)(B) subset T(B) + S(B)$, because the first depends on adding only the points of the image that come from the same vector (i.e. $T(v) + S(v)$, but not $T(v) + S(w)$ if $v neq w$) and the latter adds using all vectors in $B$ for arguments of $T$ and any vector in $B$ for an argument for $S$. Then we see that if $t$ is the smallest radius containing $T(B)$ and $s$ is the smallest radius contain $S(B)$, then the smallest radius containing $T(B) + S(B)$ is at most $t+s$, because $tB + sB = (t+s)B$, as sets.
answered Aug 23 at 7:32
4-ier
6089
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Take the (closed) unit ball $B$. It is all the points of norm at most $1$.
Now, apply the operator $T$ to the unit ball, to get a convex set $T(B)$. The operator norm of $T$ is the radius of the smallest ball centered at $0$ that will contain $T(B)$. In finite dimensions, you think of $T$ turning the unit disk into some sort of ellipse-like region and the maximal distance from the origin of the boundary of this region is the operator norm.
For example, if $|T| = 0$, then that means that $T$ sends $B$ to the single point $0$, because the smallest closed ball containing $T(B)$ is in fact just the set $0$. We then have by linearity that $T$ is identically zero.
This is why the inequality $|TS| leq |T||S|$ is actually intuitive. Let $r$ be the radius of the smallest closed ball containing $TS(B)$ (i.e. the operator norm of $TS$). If we look at the smallest radius that contains $S(B)$, $s$, and apply $T$ to that radius ball and ask what is the smallest ball containing that set, get $|T||S|$, but since this set clearly contains $TS(B)$, because instead of $T(S(B))$, we took $T(sB)$ and $S(B) subset sB$, by definition of $s$.
Likewise, you can check geometrically the triangle inequality: $|T + S| leq |T| + |S|$, because $(T + S)(B) subset T(B) + S(B)$, because the first depends on adding only the points of the image that come from the same vector (i.e. $T(v) + S(v)$, but not $T(v) + S(w)$ if $v neq w$) and the latter adds using all vectors in $B$ for arguments of $T$ and any vector in $B$ for an argument for $S$. Then we see that if $t$ is the smallest radius containing $T(B)$ and $s$ is the smallest radius contain $S(B)$, then the smallest radius containing $T(B) + S(B)$ is at most $t+s$, because $tB + sB = (t+s)B$, as sets.
answered Aug 23 at 7:32
4-ier
6089
Take the (closed) unit ball $B$. It is all the points of norm at most $1$.
Now, apply the operator $T$ to the unit ball, to get a convex set $T(B)$. The operator norm of $T$ is the radius of the smallest ball centered at $0$ that will contain $T(B)$. In finite dimensions, you think of $T$ turning the unit disk into some sort of ellipse-like region and the maximal distance from the origin of the boundary of this region is the operator norm.
For example, if $|T| = 0$, then that means that $T$ sends $B$ to the single point $0$, because the smallest closed ball containing $T(B)$ is in fact just the set $0$. We then have by linearity that $T$ is identically zero.
This is why the inequality $|TS| leq |T||S|$ is actually intuitive. Let $r$ be the radius of the smallest closed ball containing $TS(B)$ (i.e. the operator norm of $TS$). If we look at the smallest radius that contains $S(B)$, $s$, and apply $T$ to that radius ball and ask what is the smallest ball containing that set, get $|T||S|$, but since this set clearly contains $TS(B)$, because instead of $T(S(B))$, we took $T(sB)$ and $S(B) subset sB$, by definition of $s$.
Likewise, you can check geometrically the triangle inequality: $|T + S| leq |T| + |S|$, because $(T + S)(B) subset T(B) + S(B)$, because the first depends on adding only the points of the image that come from the same vector (i.e. $T(v) + S(v)$, but not $T(v) + S(w)$ if $v neq w$) and the latter adds using all vectors in $B$ for arguments of $T$ and any vector in $B$ for an argument for $S$. Then we see that if $t$ is the smallest radius containing $T(B)$ and $s$ is the smallest radius contain $S(B)$, then the smallest radius containing $T(B) + S(B)$ is at most $t+s$, because $tB + sB = (t+s)B$, as sets.
answered Aug 23 at 7:32
4-ier
6089
answered Aug 23 at 7:32
4-ier
6089
answered Aug 23 at 7:32
4-ier
6089
answered Aug 23 at 7:32
4-ier
6089
6089
add a comment |Â
add a comment |Â
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