## Unit2.3.2Geometric interpretation

We will now illustrate what the SVD Theorem tells us about matrix-vector multiplication (linear transformations) by examining the case where $A \in \R^{2 \times 2} \text{.}$ Let $A = U \Sigma V^T$ be its SVD. (Notice that all matrices are now real valued, and hence $V^H = V^T \text{.}$) Partition

\begin{equation*} A = \left( \begin{array}{c | c} u_0 \amp u_1 \end{array} \right) \left( \begin{array}{c | c} \sigma_0 \amp 0 \\ \hline 0 \amp \sigma_1 \end{array} \right) \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right)^T. \end{equation*}

Since $U$ and $V$ are unitary matrices, $\{ u_0, u_1 \}$ and $\{ v_0, v_1 \}$ form orthonormal bases for the range and domain of $A \text{,}$ respectively:

$\R^{2} \text{:}$ Domain of $A \text{:}$

$\R^{2} \text{:}$ Range (codomain) of $A \text{:}$

Let us manipulate the decomposition a little:

\begin{equation*} \begin{array}{rcl} A \amp=\amp \left( \begin{array}{c | c} u_0 \amp u_1 \end{array} \right) \left( \begin{array}{c | c} \sigma_0 \amp 0 \\ \hline 0 \amp \sigma_1 \end{array} \right) \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right)^T \\ \amp = \amp \left[ \left( \begin{array}{c | c} u_0 \amp u_1 \end{array} \right) \left( \begin{array}{c | c} \sigma_0 \amp 0 \\ \hline 0 \amp \sigma_1 \end{array} \right) \right] \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right)^T \\ \amp=\amp \left( \begin{array}{c | c} \sigma_0 u_0 \amp \sigma_1 u_1 \end{array} \right) \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right)^T. \end{array} \end{equation*}

Now let us look at how $A$ transforms $v_0$ and $v_1 \text{:}$

\begin{equation*} A v_0 = \left( \begin{array}{c | c} \sigma_0 u_0 \amp \sigma_1 u_1 \end{array} \right) \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right)^T v_0 = \left( \begin{array}{c | c} \sigma_0 u_0 \amp \sigma_1 u_1 \end{array} \right) \left( \begin{array}{c} 1 \\ \hline 0 \end{array} \right) = \sigma_0 u_0 \end{equation*}

and similarly $A v_1 = \sigma_1 u_1 \text{.}$ This motivates the pictures in FigureĀ 2.3.2.1.

Next, let us look at how $A$ transforms any vector with (Euclidean) unit length. Notice that $x = \left( \begin{array}{c} \chi_0 \\ \chi_1 \end{array} \right)$ means that

\begin{equation*} x = \chi_0 e_0 + \chi_1 e_1 , \end{equation*}

where $e_0$ and $e_1$ are the unit basis vectors. Thus, $\chi_0$ and $\chi_1$ are the coefficients when $x$ is expressed using $e_0$ and $e_1$ as basis. However, we can also express $x$ in the basis given by $v_0$ and $v_1 \text{:}$

\begin{equation*} \begin{array}{rcl} x \amp=\amp \begin{array}[t]{c} \underbrace{V V^T} \\ I \end{array} x = \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right) \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right)^T x = \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right) \left( \begin{array}{c} v_0^T x \\ \hline v_1^T x \end{array} \right)\\ \amp=\amp \begin{array}[t]{c} \underbrace{ v_0^T x } \\ \alpha_0 \end{array} v_0 + \begin{array}[t]{c} \underbrace{ v_1^T x } \\ \alpha_1 \end{array} v_1 = \alpha_0 v_0 + \alpha_1 v_1 = \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right) \left( \begin{array}{c} \alpha_0 \\ \alpha_1 \end{array} \right) . \end{array} \end{equation*}

Thus, in the basis formed by $v_0$ and $v_1 \text{,}$ its coefficients are $\alpha_0$ and $\alpha_1 \text{.}$ Now,

\begin{equation*} \begin{array}{rcl} A x \amp=\amp \left( \begin{array}{c | c} \sigma_0 u_0 \amp \sigma_1 u_1 \end{array} \right) \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right)^T x \\ \amp = \amp \left( \begin{array}{c | c} \sigma_0 u_0 \amp \sigma_1 u_1 \end{array} \right) \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right)^T \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right) \left( \begin{array}{c} \alpha_0 \\ \alpha_1 \end{array} \right) \\ \amp=\amp \left( \begin{array}{c | c} \sigma_0 u_0 \amp \sigma_1 u_1 \end{array} \right) \left( \begin{array}{c} \alpha_0 \\ \alpha_1 \end{array} \right) = \alpha_0 \sigma_0 u_0 + \alpha_1 \sigma_1 u_1. \end{array} \end{equation*}

This is illustrated by the following picture, which also captures the fact that the unit ball is mapped to an oval with major axis equal to $\sigma_0 = \| A \|_2$ and minor axis equal to $\sigma_1 \text{,}$ as illustrated in FigureĀ 2.3.2.1 (bottom).

Finally, we show the same insights for general vector $x$ (not necessarily of unit length):

$\R^{2} \text{:}$ Domain of $A \text{:}$

$\R^{2} \text{:}$ Range (codomain) of $A \text{:}$

Another observation is that if one picks the right basis for the domain and codomain, then the computation $A x$ simplifies to a matrix multiplication with a diagonal matrix. Let us again illustrate this for nonsingular $A \in \R^{2 \times 2}$ with

\begin{equation*} A = \begin{array}[t]{c} \underbrace{ \left( \begin{array}{c | c} u_0 \amp u_1 \end{array} \right) } \\ U \end{array} \begin{array}[t]{c} \underbrace{ \left( \begin{array}{c | c} \sigma_0 \amp 0 \\ \hline 0 \amp \sigma_1 \end{array} \right) } \\ \Sigma \end{array} \begin{array}[t]{c} \underbrace{ \left( \begin{array}{c | c} v_0 \amp v_1 \end{array} \right) } \\ V \end{array} ^T. \end{equation*}

Now, if we chose to express $y$ using $u_0$ and $u_1$ as the basis and express $x$ using $v_0$ and $v_1$ as the basis, then

\begin{equation*} \begin{array}{rcl} \begin{array}[t]{c} \underbrace{U U^T} \\ I \end{array} y \amp=\amp U \begin{array}[t]{c} \underbrace{U^T y} \\ \widehat y \end{array} = ( u_0^T y ) u_0 + ( u_1^T y ) u_1 \\ \amp=\amp \left(\begin{array}{c | c} u_0 \amp u_1 \end{array} \right) \left( \begin{array}{c} u_0^T y \\ \hline u_1^T y \end{array}\right) = U \begin{array}[t]{c} \underbrace{ \left( \begin{array}{c} \widehat \psi_0 \\ \hline \widehat \psi_1 \end{array} \right)}\\ \widehat y \end{array} \\ \begin{array}[t]{c} \underbrace{V V^T} \\ I \end{array} x \amp=\amp V \begin{array}[t]{c} \underbrace{V^T x} \\ \widehat x \end{array} = ( v_0^T x ) v_0 + ( v_1^T x ) v_1 \\ \amp=\amp \left(\begin{array}{c | c} v_0 \amp v_1 \end{array} \right) \left( \begin{array}{c} v_0^T x \\ \hline v_1^T x \end{array}\right) = V \begin{array}[t]{c} \underbrace{ \left( \begin{array}{c} \widehat \chi_0 \\ \hline \widehat \chi_1. \end{array} \right)} \\ \widehat x \end{array}. \end{array} \end{equation*}

If $y = A x$ then

\begin{equation*} U \begin{array}[t]{c} \underbrace{ U^T y } \\ \widehat y \end{array} = \begin{array}[t]{c} \underbrace{ U \Sigma V^T x } \\ A x \end{array} = U \Sigma \widehat x \end{equation*}

so that

\begin{equation*} \widehat y = \Sigma \widehat x \end{equation*}

and

\begin{equation*} \left( \begin{array}{c} \widehat \psi_0 \\ \hline \widehat \psi_1. \end{array} \right) = \left( \begin{array}{c} \sigma_0 \widehat \chi_0 \\ \hline \sigma_1 \widehat \chi_1. \end{array} \right). \end{equation*}
###### Remark2.3.2.2.

The above discussion shows that if one transforms the input vector $x$ and output vector $y$ into the right bases, then the computation $y := A x$ can be computed with a diagonal matrix instead: $\widehat y := \Sigma \widehat x \text{.}$ Also, solving $A x = y$ for $x$ can be computed by multiplying with the inverse of the diagonal matrix: $\widehat x := \Sigma^{-1} \widehat y \text{.}$

These observations generalize to $A \in \C^{m \times n}\text{:}$ If

\begin{equation*} y = A x \end{equation*}

then

\begin{equation*} U^H y = U^H A \begin{array}[t]{c} \underbrace{V V^H }\\ I \end{array} x \end{equation*}

so that

\begin{equation*} \begin{array}[t]{c} \underbrace{U^H y}\\ \widehat y \end{array} = \Sigma \begin{array}[t]{c} \underbrace{V^H x}\\ \widehat x \end{array} \end{equation*}

($\Sigma$ is a rectangular "diagonal" matrix.)