91Ó°ÊÓ

Consider an SVD

$A=U\underset{}{\overset{}{∑V^T}}$

of an$n×mmatrixA$. If U is an orthogonal$n×n$ matrix, then V is an orthogonal $m×m$ matrix, and$\underset{}{\overset{}{∑}}$ is an$n×m$ matrix whose first r diagonal entries are the nonzero singular values${\mathrm{σ}}_1,{\mathrm{σ}}_2,......,{\mathrm{σ}}_{r}ofA$ , and all other entries are zero. Thus, the rank of A will be r. Therefore, we have

$A={\mathrm{σ}}_1{\stackrel{→}{u}}_1{\stackrel{→}{v}}_1^T+{\mathrm{σ}}_2{\stackrel{→}{u}}_2{\stackrel{→}{v}}_2^T+.........+{\mathrm{σ}}_r{\stackrel{→}{u}}_r{\stackrel{→}{v}}_r^T$

where ${\stackrel{→}{u}}_{i}and{\stackrel{→}{v}}_i$ are the columns of U and V respectively. Since $A=U\underset{}{\overset{}{∑V^T}}$

we get

$A^T=V\underset{}{\overset{}{{∑}^T{U}^T}}$

We also have

$$\begin{gathered} A{\stackrel{→}{v}}_i={\mathrm{σ}}_i{\stackrel{→}{u}}_i,(fori=1,.......,r) \\ A{\stackrel{→}{v}}_i=0(fori=r+1,.....,n) \\ {A}^T{\stackrel{→}{u}}_i\&={\mathrm{σ}}_i{\stackrel{→}{v}}_i(fori=1......r) \\ {A}^T{\stackrel{→}{u}}_i=0(fori=r+1.....n) \end{gathered}$$

Hence we get as follows:

$$\begin{gathered} A^T{\stackrel{→}{u}}_i\&={\mathrm{σ}}_i{\stackrel{→}{v}}_i \\ ⇒A{A}^T{\stackrel{→}{u}}_i=A\left({\mathrm{σ}}_i{\stackrel{→}{v}}_i\right) \\ ⇒A{A}^T{\stackrel{→}{u}}_i={\mathrm{σ}}_{i}A{\stackrel{→}{v}}_i \\ ⇒A{A}^T{\stackrel{→}{u}}_i=\left\{\begin{array}{ll}{\mathrm{σ}}_i^2{\stackrel{→}{u}}_{i}fori=1.........,r& (\sin ceA{\stackrel{→}{v}}_i={\mathrm{σ}}_i{\stackrel{→}{u}}_i\\ 0& fori=r+1......n(\sin ceA{\stackrel{→}{v}}_i=0)\end{array}\right. \end{gathered}$$

Therefore, we can infer from the above equation that the non zero eigen values of$A^{T}AandA{A}^T$ are the same.

Let A be an $n×m$matrix.

Since U is an orthogonal matrix, hence the nonzero columns of U form an orthonormal set Also, let u be a nonzero column of U. then the corresponding singular value role="math" localid="1659687933192" $\mathrm{σ}ofA$is also non-zero. Hence ${\mathrm{σ}}^2$is a nonzero eigenvalue of $A^{t}A$Now by the construction of U, we have

$u=\frac{1}{\mathrm{σ}}Av$

where is a eigenvector of corresponding to the eigenvalue . Then we have

$$\begin{gathered} \left(A{A}^t\right)u=\left(A{A}^t\right)\left(\frac{1}{\mathrm{σ}}Av\right) \\ =\frac{1}{\mathrm{σ}}\left(\left(A{A}^t\right)Av\right) \\ =\frac{1}{\mathrm{σ}}\left(A\left(A{A}^t\right)v\right) \\ =\frac{1}{\mathrm{σ}}\left(A({\mathrm{σ}}^{2}v\right)\left[\sin cevisaeigenvectorof{A}^{t}Acorrespondingtotheeignvalue{\mathrm{σ}}^2\right] \\ =\mathrm{σ}\left(Av\right) \\ =\mathrm{σ}\left(\mathrm{σ}v\right)\left[\sin ceu=\frac{1}{\mathrm{σ}}Av\right] \\ ={\mathrm{σ}}^{2}u \end{gathered}$$

This shows that u is a eigenvector of $A{A}^t$corresponding to the eigenvalue ${\mathrm{σ}}^2$

This shows that the nonzero columns of U form an orthonormal eigenbasis for $A{A}^t$ The associated eigenvalues are ${\mathrm{σ}}^{2}where\mathrm{σ}isa\sin gularvaluesofA$

This shows that the nonzero eigenvalues of$A{A}^{t}and{A}^{t}A$ are same.

Also, if 0 is an eigenvalue ofrole="math" localid="1659689046090" $A{A}^t,thendet\left(A{A}^t\right)=0$ which implies that$det\left(A\right)=0$ Hence$det\left(A^{t}A\right)=0$ .

Thus we get that 0 is an eigenvalue of $A^{t}A$as well.

Hence all the eigenvalues role="math" localid="1659688967449" $A{A}^{t}and{A}^{t}A$of are same.

all the eigenvalues$A{A}^{t}and{A}^{t}A$ of are same.