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Q7.4-19E

Expert-verified
Found in: Page 395

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# In Exercises 17–24, $$A$$ is an $$m \times n$$ matrix with a singular value decomposition $$A = U\Sigma {V^T}$$ , where $$U$$ is an $$m \times m$$ orthogonal matrix, $${\bf{\Sigma }}$$ is an $$m \times n$$ “diagonal” matrix with $$r$$ positive entries and no negative entries, and $$V$$ is an $$n \times n$$ orthogonal matrix. Justify each answer.19. Show that the columns of $$V$$ are eigenvectors of $${A^T}A$$, the columns of $$U$$ are eigenvectors of $$A{A^T}$$, and the diagonal entries of $${\bf{\Sigma }}$$ are the singular values of $$A$$. (Hint: Use the SVD to compute $${A^T}A$$ and $$A{A^T}$$.)

It is verified that, $$V$$ and $$U$$ are eigenvectors of $${A^T}A$$ and $$A{A^T}$$ respectively, and the columns of $$U$$ re eigenvectors of $$A{A^T}$$ and the diagonal elements of $$\Sigma$$ are the singular values of $$A$$.

See the step by step solution

## Step 1: Find the product of $${A^T}$$ and $$A$$

As singular value decomposition of $$A$$ is $$A = U\Sigma {V^T}$$, then

$$\begin{array}{c}{A^T}A = {\left( {U\Sigma {V^T}} \right)^T}U\Sigma {V^T}\\ = V\mathop \Sigma \limits^T {U^T}\Sigma {V^T}\\ = V\left( {\mathop \Sigma \limits^T \Sigma } \right){V^T}\\ = V\left( {\mathop \Sigma \limits^T \Sigma {V^{ - 1}}} \right)\end{array}$$

Therefore, from the Diagonalization theorem. Singular values are the diagonal entries in $$\Sigma$$ matrix. The columns of matrix $$V$$ are the eigenvectors of $${A^T}A$$.

## Step 2: Find the product of $$A$$ and $${A^T}$$

Now, find the product of $$A = U\Sigma {V^T}$$ and its transpose.

$$\begin{array}{c}A{A^T} = U\Sigma {V^T}{\left( {U\Sigma {V^T}} \right)^T}\\ = U\mathop \Sigma \limits^T {V^T}V\mathop \Sigma \limits^T {U^T}\\ = U\left( {\Sigma \mathop \Sigma \limits^T } \right){U^T}\\ = U\left( {\Sigma \mathop \Sigma \limits^T } \right){U^{ - 1}}\end{array}$$

Thus, it can be observed that $$U$$ diagonalizes $$A{A^T}$$ and the columns of $$U$$ must be eigenvectors of $$A{A^T}$$.