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Q7.4-26E

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

# (M) Compute an SVD of each matrix in Exercises 26 and 27. Report the final matrix entries accurate to two decimal places. Use the method of Examples 3 and 4.26. $$A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{ - {\bf{18}}}&{{\bf{13}}}&{ - {\bf{4}}}&{\bf{4}}\\{\bf{2}}&{{\bf{19}}}&{ - {\bf{4}}}&{{\bf{12}}}\\{ - {\bf{14}}}&{{\bf{11}}}&{ - {\bf{12}}}&{\bf{8}}\\{ - {\bf{2}}}&{{\bf{21}}}&{\bf{4}}&{\bf{8}}\end{array}} \right)$$

The singular value decomposition of $$A$$ is:$$A = \left( {\begin{array}{*{20}{c}}{.5}&{ - .5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{ - .5}\\{.5}&{ - .5}&{.5}&{.5}\\{.5}&{.5}&{ - .5}&{.5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{40}&0&0&0\\0&{20}&0&0\\0&0&{10}&0\\0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - .4}&{.8}&{ - .2}&{.4}\\{.8}&{.4}&{.4}&{.2}\\{.4}&{ - .2}&{ - .8}&{.4}\\{ - .2}&{ - .4}&{.4}&{.8}\end{array}} \right)$$

See the step by step solution

## Step 1: Write the matrix

Find the transpose of the matrix$$A$$ by using MATLAB commands.

$$> > B = A';$$

$$B = \left( {\begin{array}{*{20}{c}}{ - 18}&2&{ - 14}&{ - 2}\\{13}&{19}&{11}&{21}\\{ - 4}&{ - 4}&{ - 12}&4\\4&{12}&8&8\end{array}} \right)$$

Find the product of $$A$$ and $$B$$.

$$> > {\rm{ }}C = B*A;$$

$$C = \left( {\begin{array}{*{20}{c}}{528}&{ - 392}&{224}&{ - 176}\\{ - 392}&{1092}&{ - 176}&{536}\\{224}&{ - 176}&{192}&{ - 128}\\{ - 176}&{536}&{ - 128}&{288}\end{array}} \right)$$

## Step 2: Find eigenvalues and eigenvectors of the matrix $$AB$$

Find the eigenvectors of the matrix $$AB$$.

$$> > {\rm{ }}\left( {V{\rm{ }}B} \right) = eigs\left( C \right);$$

$$V = \left( {\begin{array}{*{20}{c}}{ - \frac{2}{5}}&{\frac{4}{5}}&{\frac{2}{5}}&{ - \frac{1}{5}}\\{\frac{4}{5}}&{\frac{2}{5}}&{ - \frac{1}{5}}&{ - \frac{2}{5}}\\{ - \frac{1}{5}}&{\frac{2}{5}}&{ - \frac{4}{5}}&{\frac{2}{5}}\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{2}{5}}&{\frac{2}{5}}\end{array}} \right)$$

$$\begin{array}{l} > > {\rm{ }}SIGMA = {\rm{ }}diag\left( {sqrt\left( E \right)} \right);\\ > > {\rm{ }}SS = sqrt\left( E \right);\end{array}$$

Then we have $$\sum$$ the matrix.

$$\sum = \left( {\begin{array}{*{20}{c}}{40}&0&0&0\\0&{20}&0&0\\0&0&{10}&0\\0&0&0&0\end{array}} \right)$$

## Step 3: Find the singular value decomposition of$$A$$

Write the vectors of the $$U$$ matrix:

$$\begin{array}{l} > > {\rm{ }}u1{\rm{ }} = {\rm{ }}\left( {1/SS\left( 1 \right)} \right)*A*V\left( {:,1} \right);\\ > > {\rm{ }}u2{\rm{ }} = {\rm{ }}\left( {1/SS\left( 2 \right)} \right)*A*V\left( {:,2} \right);\\ > > {\rm{ }}u3{\rm{ }} = {\rm{ }}\left( {1/SS\left( 3 \right)} \right)*A*V\left( {:,3} \right);\\ > > {\rm{ }}u4{\rm{ }} = {\rm{ }}(1/SS\left( 4 \right)*A*V\left( {:,4} \right);\\ > > {\rm{ }}U = \left( {u1;u2;u3;u4} \right);\end{array}$$

$$U = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\end{array}} \right)$$

Therefore, the singular value decomposition of $$A$$ is:

$$\begin{array}{c}A = U\sum {V^T}\\ = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{40}&0&0&0\\0&{20}&0&0\\0&0&{10}&0\\0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - \frac{2}{5}}&{\frac{4}{5}}&{\frac{2}{5}}&{ - \frac{1}{5}}\\{\frac{4}{5}}&{\frac{2}{5}}&{ - \frac{1}{5}}&{ - \frac{2}{5}}\\{ - \frac{1}{5}}&{\frac{2}{5}}&{ - \frac{4}{5}}&{\frac{2}{5}}\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{2}{5}}&{\frac{2}{5}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{.5}&{ - .5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{ - .5}\\{.5}&{ - .5}&{.5}&{.5}\\{.5}&{.5}&{ - .5}&{.5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{40}&0&0&0\\0&{20}&0&0\\0&0&{10}&0\\0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - .4}&{.8}&{ - .2}&{.4}\\{.8}&{.4}&{.4}&{.2}\\{.4}&{ - .2}&{ - .8}&{.4}\\{ - .2}&{ - .4}&{.4}&{.8}\end{array}} \right)\end{array}$$

Thus, $$A = \left( {\begin{array}{*{20}{c}}{.5}&{ - .5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{ - .5}\\{.5}&{ - .5}&{.5}&{.5}\\{.5}&{.5}&{ - .5}&{.5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{40}&0&0&0\\0&{20}&0&0\\0&0&{10}&0\\0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - .4}&{.8}&{ - .2}&{.4}\\{.8}&{.4}&{.4}&{.2}\\{.4}&{ - .2}&{ - .8}&{.4}\\{ - .2}&{ - .4}&{.4}&{.8}\end{array}} \right)$$.