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Subsection 4.3.1 The SVD and the four fundamental spaces

We prove that \(\Col( A ) = \Col( U_L ) \text{,}\) leaving the other parts as exercises.

Let \(A = U_L \Sigma_{TL} V_L^H \) be the Reduced SVD of \(A \text{.}\) Then

  • \(U_L^H U_L = I \) (\(U_L \) is orthonormal),

  • \(V_L^H V_L = I \) (\(V_L \) is orthonormal), and

  • \(\Sigma_{TL} \) is nonsingular because it is diagonal and the diagonal elements are all nonzero.

We will show that \(\Col(A) = \Col( U_L ) \) by showing that \(\Col(A) \subset \Col( U_L ) \) and \(\Col(U_L) \subset \Col( A ) \)

  • \(\Col(A) \subset \Col( U_L ) \text{:}\)

    Let \(z \in \Col(A) \text{.}\) Then there exists a vector \(x \in \Cn \) such that \(z = A x \text{.}\) But then \(z = A x = U_L \Sigma_{TL} V_L^H x = U_L \begin{array}[t]{c} \underbrace{\Sigma_{TL} V_L^H x}\\ \widehat x \end{array} = U_L \widehat x \text{.}\) Hence \(z \in \Col( U_L ) \text{.}\)

  • \(\Col(U_L) \subset \Col(A) \text{:}\)

    Let \(z \in \Col(U_L) \text{.}\) Then there exists a vector \(x \in \C^r \) such that \(z = U_L x \text{.}\) But then \(z = U_L x = U_L \begin{array}[t]{c} \underbrace{ \Sigma_{TL} V_L^H V_L \Sigma_{TL}^{-1}}\\ I \end{array} x = A \begin{array}[t]{c} \underbrace{ V_L \Sigma_{TL}^{-1} x}\\ \widehat x \end{array} = A \widehat x \text{.}\) Hence \(z \in \Col( A ) \text{.}\)

We leave the other parts as exercises for the learner.

Homework 4.3.1.1.

For the last theorem, prove that \(\Rowspace( A ) = \Col( A^H ) = \Col( V_L ) \text{.}\)

Solution

\(\Rowspace( A ) = \Col( V_L ) \text{:}\)

The slickest way to do this is to recognize that if \(A = U_L \Sigma_{TL} V_L^H \) is the Reduced SVD of \(A \) then \(A^H = V_L \Sigma_{TL} U_L^H \) is the Reduced SVD of \(A^H \text{.}\) One can then invoke the fact that \(\Col( A ) = \Col( U_L ) \) where in this case \(A \) is replaced by \(A^H \) and \(U_L \) by \(V_L \text{.}\)

Ponder This 4.3.1.2.

For the last theorem, prove that \(\Null( A^H ) = \Col( U_R ) \text{.}\)

Homework 4.3.1.3.

Given \(A \in \C^{m \times n}\text{,}\) let \(A = U_L \Sigma_{TL} V_L^H \) equal its Reduced SVD and \(A = \left( \begin{array}{c | c} U_L \amp U_R \end{array} \right) \left( \begin{array}{c | c} \Sigma_{TL} \amp 0 \\ \hline 0 \amp 0 \end{array} \right) \left( \begin{array}{c | c} V_L \amp V_R \end{array} \right)^H \) its SVD, and \(r = \rank( A ) \text{.}\)

  • ALWAYS/SOMETIMES/NEVER: \(r = \rank( A ) = \dim( \Col( A ) ) = \dim( \Col( U_L ) ) \text{,}\)

  • ALWAYS/SOMETIMES/NEVER: \(r = \dim( \Rowspace( A ) ) = \dim( \Col( V_L ) ) \text{,}\)

  • ALWAYS/SOMETIMES/NEVER: \(n-r = \dim( \Null( A ) ) = \dim( \Col( V_R) ) \text{,}\) and

  • ALWAYS/SOMETIMES/NEVER: \(m-r = \dim( \Null( A^H ) ) = \dim( \Col( U_R ) ) \text{.}\)

Answer

  • ALWAYS: \(r = \rank( A ) = \dim( \Col( A ) ) = \dim( \Col( U_L ) ) \text{,}\)

  • ALWAYS: \(r = \dim( \Rowspace( A ) ) = \dim( \Col( V_L ) ) \text{,}\)

  • ALWAYS: \(n-r = \dim( \Null( A ) ) = \dim( \Col( V_R) ) \text{,}\) and

  • ALWAYS: \(m-r = \dim( \Null( A^H ) ) = \dim( \Col( U_R ) ) \text{.}\)

Now prove it.

Solution

  • ALWAYS: \(r = \rank( A ) = \dim( \Col( A ) ) = \dim( \Col( U_L ) ) \text{,}\)

    The dimension of a space equals the number of vectors in a basis. A basis is any set of linearly independent vectors such that the entire set can be created by taking linear combinations of those vectors. The rank of a matrix is equal to the dimension of its columns space which is equal to the dimension of its row space.

    Now, clearly the columns of \(U_L \) are linearly independent (since they are orthonormal) and form a basis for \(\Col( U_L ) \text{.}\) This, together with Theorem 4.3.1.1, yields the fact that \(r = \rank( A ) = \dim( \Col( A ) ) = \dim( \Col( U_L ) ) \text{.}\)

  • ALWAYS: \(r = \dim( \Rowspace( A ) ) = \dim( \Col( V_L ) ) \text{,}\)

    There are a number of ways of reasoning this. One is a small modification of the proof that \(r = \rank( A ) = \dim( \Col( A ) ) = \dim( \Col( U_L ) ) \text{.}\) Another is to look at \(A^H \) and to apply the last subproblem.

  • ALWAYS: \(n-r = \dim( \Null( A ) ) = \dim( \Col( V_R) ) \text{.}\)

    We know that \(\dim( \Null( A ) ) + \dim( \Rowspace( A ) ) = n \text{.}\) The answer follows directly from this and the last subproblem.

  • ALWAYS: \(m-r = \dim( \Null( A^H ) ) = \dim( \Col( U_R ) ) \text{.}\)

    We know that \(\dim( \Null( A^H ) ) + \dim( \Col( A ) ) = m \text{.}\) The answer follows directly from this and the first subproblem.

Homework 4.3.1.4.

Given \(A \in \C^{m \times n}\text{,}\) let \(A = U_L \Sigma_{TL} V_L^H \) equal its Reduced SVD and \(A = \left( \begin{array}{c | c} U_L \amp U_R \end{array} \right) \left( \begin{array}{c | c} \Sigma_{TL} \amp 0 \\ \hline 0 \amp 0 \end{array} \right) \left( \begin{array}{c | c} V_L \amp V_R \end{array} \right)^H \) its SVD.

Any vector \(x \in \Cn \) can be written as \(x = x_r + x_n \) where \(x_r \in \Col( V_L ) \) and \(x_n \in \Col( V_R ) \text{.}\)

TRUE/FALSE

Answer

TRUE

Now prove it!

Solution
\begin{equation*} \begin{array}{l} x = I x = V V^H x \\ ~~~= \left( \begin{array}{c | c} V_L \amp V_R \end{array} \right) \left( \begin{array}{c | c} V_L \amp V_R \end{array} \right)^H x \\ ~~~= \left( \begin{array}{c | c} V_L \amp V_R \end{array} \right) \left( \begin{array}{c} V_L^H \\ V_R^H \end{array} \right) x \\ ~~~= \left( \begin{array}{c | c} V_L \amp V_R \end{array} \right) \left( \begin{array}{c} V_L^H x \\ V_R^H x\end{array} \right) \\ ~~~= \begin{array}[t]{c} \underbrace{ V_L V_L^H x } \\ x_r \end{array} + \begin{array}[t]{c} \underbrace{ V_R V_R^H x } \\ x_n \end{array} . \end{array} \end{equation*}
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Figure 4.3.1.2. Illustration of relationship between the SVD of matrix \(A \) and the four fundamental spaces.
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