2.4 Exercises

20 points · 4 extra credit

Partitioned Matrices

1. 1. (3 pts) Compute the block-diagonal product $\begin{bmatrix} A_1 & 0 \\ 0 & A_2 \end{bmatrix}\begin{bmatrix} B_1 & 0 \\ 0 & B_2 \end{bmatrix}$ (each block $2\times2$), where

   $A_1 = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix},\; A_2 = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix},\; B_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},\; B_2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}.$

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2. 2. (3 pts) Find the inverse of the block-diagonal matrix $M = \begin{bmatrix} A & 0 \\ 0 & D \end{bmatrix}$, where $A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$ and $D = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.

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3. 3. (3 pts) Let $A$ and $D$ be invertible. Verify by direct block multiplication that

   $\begin{bmatrix} A & 0 \\ C & D \end{bmatrix}^{-1} = \begin{bmatrix} A^{-1} & 0 \\ -D^{-1}CA^{-1} & D^{-1} \end{bmatrix}.$

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4. 4. (3 pts) Let $P = \begin{bmatrix} I & 0 \\ C & I \end{bmatrix}$ and $M = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}$ with conformable blocks. Compute the product $PM$ in block form, and describe in words the "block row operation" that $P$ performs on $M$.

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5. 5. (3 pts) State whether each is true or false, and justify.

   (a) If $A$ and $B$ are partitioned conformably for block multiplication, then the product $AB$ can be computed as though the blocks were scalars.

   (b) The transpose of $\begin{bmatrix} A & B \\ C & D \end{bmatrix}$ is $\begin{bmatrix} A^{\mathsf T} & B^{\mathsf T} \\ C^{\mathsf T} & D^{\mathsf T} \end{bmatrix}$.

   (c) A block-diagonal matrix $\operatorname{diag}(A_1, \ldots, A_k)$ is invertible if and only if every diagonal block $A_i$ is invertible.

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6. 6. (3 pts) Using the column–row expansion of a product, write $AB$ as a sum of outer products (column of $A$ times row of $B$), and evaluate for

   $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \qquad B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}.$

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7. 7. (2 pts) Explain why, for the block product $\begin{bmatrix} A_{11} & A_{12} \end{bmatrix}\begin{bmatrix} B_{11} \\ B_{21} \end{bmatrix} = A_{11}B_{11} + A_{12}B_{21}$ to make sense, the column partition of the left factor must match the row partition of the right factor.

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8. 8. (4 pts ••) Let $M = \begin{bmatrix} A & B \\ C & D \end{bmatrix}$ with $A$ invertible, and let $S = D - CA^{-1}B$ (the Schur complement of $A$) be invertible. Verify the block-LU factorization

   $M = \begin{bmatrix} I & 0 \\ CA^{-1} & I \end{bmatrix}\begin{bmatrix} A & B \\ 0 & S \end{bmatrix},$

   and use it to explain why $M$ is invertible.

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