### Linear Algebra for Graphics Geeks (SVD-VI)

This is post #6 of Linear Algebra for Graphics Geeks. Here are links to posts #1, #2, #3, #4, #5. The subject at hand is the Singular Value Decomposition.

This time I'm going to discuss eigenvectors, eigenvalues and ways in which they are related to the SVD.

For all the complaints levied against Wikipedia, I am often pleased by how clearly and succinctly many subjects are explained:

"An eigenvector of a transformation[1] is a non-null vector whose direction is unchanged by that transformation. The factor by which the magnitude is scaled is called the eigenvalue of that vector"

Let's say you have a matrix A and a vector V and you transform V with A. If A has no effect on the direction of V and all that A does is make V longer or shorter (or the same), then V is an eigenvector of A:

A*V = scale*V

The value of "scale" is the eigenvalue for the eigenvector V.

Take our matrix that scales X by 3 and Y by 2:

[3 0]

[0 2]

The eigenvectors of this matrix are [1 0]' and [0 1]'. The eigenvalues are 3 and 2.

[3 0] * [1] = [3] = 3 * [1]

[0 2] . [0] . [0] . . . [0]

[3 0] * [0] = [0] = 2 * [0]

[0 2] . [1] . [2] . . . [1]

Transformation of the unit vector on the x axis [1 0]' results in [3 0]', a 3X scaling of its length. Likewise, transformation of the unit vector on the y axis [0 1]' results in [0 2]', a 2X scaling of its length.

The eigen decomposition takes the following form:

A = P*D*P

Where P is an orthogonal matrix and D is a diagonal matrix.

This decomposition can always be done on a symmetric matrix (Spectral Theorem). Since the product of a matrix and its transpose is always a symmetric matrix, this decomposition can always be done on A*A' and A'*A.

While we're on the subject, it's interesting to consider the the square of a symmetric matrix given its eigen decomposition.

A

= P*D*P

= P*D*(P

= P*D*I*D*P

= P*D*D*P

= P*D

Note that squaring the matrix results in a squaring of D, and since D is a diagonal matrix, this amounts to a squaring of each element on D's diagonal.

We've gotten far enough to note more interesting properties of SVD:

1. The columns of U are the eigenvectors of A*A'

2. The columns of V are the eigenvectors of A'*A

3. The diagonal elements of S are the square roots of the eigenvalues of A*A' and A'*A.

Like many other proporties of the SVD laid out so far, these three facts also have profound implications. Hopefully, I'll have time to look into it much more deeply in a future post, but I think everything above is more than enough to ponder for now.

I'll leave with some equations that get us from the SVD to the eigen decompositions of A*A' and A'*A.

A = U*S*V'

A' = (U*S*V')'

A' = V*S*U'

A*A' = (U*S*V')*(V*S*U')

= U*S*V'*V*S*U'

= U*S*(V'*V)*S*U'

= U*S*I*S*U'

= U*S

A'*A = (V*S*U')*(U*S*V') =

V*S*U'*U*S*V' =

V*S*(U'*U)*S*V' =

V*S*I*S*V' =

V*S

In part #7, I look at the SVD of a singular matrix.

This time I'm going to discuss eigenvectors, eigenvalues and ways in which they are related to the SVD.

For all the complaints levied against Wikipedia, I am often pleased by how clearly and succinctly many subjects are explained:

"An eigenvector of a transformation[1] is a non-null vector whose direction is unchanged by that transformation. The factor by which the magnitude is scaled is called the eigenvalue of that vector"

Let's say you have a matrix A and a vector V and you transform V with A. If A has no effect on the direction of V and all that A does is make V longer or shorter (or the same), then V is an eigenvector of A:

A*V = scale*V

The value of "scale" is the eigenvalue for the eigenvector V.

*Eigenvalues are usually identified with the symbol lamba.*Take our matrix that scales X by 3 and Y by 2:

[3 0]

[0 2]

The eigenvectors of this matrix are [1 0]' and [0 1]'. The eigenvalues are 3 and 2.

[3 0] * [1] = [3] = 3 * [1]

[0 2] . [0] . [0] . . . [0]

[3 0] * [0] = [0] = 2 * [0]

[0 2] . [1] . [2] . . . [1]

Transformation of the unit vector on the x axis [1 0]' results in [3 0]', a 3X scaling of its length. Likewise, transformation of the unit vector on the y axis [0 1]' results in [0 2]', a 2X scaling of its length.

The eigen decomposition takes the following form:

A = P*D*P

^{-1}Where P is an orthogonal matrix and D is a diagonal matrix.

This decomposition can always be done on a symmetric matrix (Spectral Theorem). Since the product of a matrix and its transpose is always a symmetric matrix, this decomposition can always be done on A*A' and A'*A.

*Note to self: In a revision, I'd like to say something about the relationship between the eigen decomposition and the ellipse equation. One form of the equation for the ellipse is ax*

^{2}+ bx^{2}= 1.While we're on the subject, it's interesting to consider the the square of a symmetric matrix given its eigen decomposition.

A

^{2}= A*A = (P*D*P^{-1})*(P*D*P^{-1})= P*D*P

^{-1}*P*D*P^{-1}= P*D*(P

^{-1}*P)*D*P^{-1}= P*D*I*D*P

^{-1}= P*D*D*P

^{-1}= P*D

^{2}*P^{-1}Note that squaring the matrix results in a squaring of D, and since D is a diagonal matrix, this amounts to a squaring of each element on D's diagonal.

We've gotten far enough to note more interesting properties of SVD:

1. The columns of U are the eigenvectors of A*A'

2. The columns of V are the eigenvectors of A'*A

3. The diagonal elements of S are the square roots of the eigenvalues of A*A' and A'*A.

Like many other proporties of the SVD laid out so far, these three facts also have profound implications. Hopefully, I'll have time to look into it much more deeply in a future post, but I think everything above is more than enough to ponder for now.

I'll leave with some equations that get us from the SVD to the eigen decompositions of A*A' and A'*A.

A = U*S*V'

A' = (U*S*V')'

A' = V*S*U'

A*A' = (U*S*V')*(V*S*U')

= U*S*V'*V*S*U'

= U*S*(V'*V)*S*U'

= U*S*I*S*U'

= U*S

^{2}*U'A'*A = (V*S*U')*(U*S*V') =

V*S*U'*U*S*V' =

V*S*(U'*U)*S*V' =

V*S*I*S*V' =

V*S

^{2}* V'*Note: This is true when A is rectangular as well as square.*In part #7, I look at the SVD of a singular matrix.

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