The Timpson Lyceum seal

The Timpson Lyceum · Linear Algebra (2026)

2.4 Exercises

Problem sheet · 20 points · 4 extra credit

Partitioned Matrices

  1. 1. (3 pts) Compute the block-diagonal product [A100A2][B100B2]\begin{bmatrix} A_1 & 0 \\ 0 & A_2 \end{bmatrix}\begin{bmatrix} B_1 & 0 \\ 0 & B_2 \end{bmatrix} (each block 2×22\times2), where

    A1=[1234],  A2=[2011],  B1=[0110],  B2=[1201].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}.

  2. 2. (3 pts) Find the inverse of the block-diagonal matrix M=[A00D]M = \begin{bmatrix} A & 0 \\ 0 & D \end{bmatrix}, where A=[2003]A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} and D=[1101]D = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}.

  3. 3. (3 pts) Let AA and DD be invertible. Verify by direct block multiplication that

    [A0CD]−1=[A−10−D−1CA−1D−1].\begin{bmatrix} A & 0 \\ C & D \end{bmatrix}^{-1} = \begin{bmatrix} A^{-1} & 0 \\ -D^{-1}CA^{-1} & D^{-1} \end{bmatrix}.

  4. 4. (3 pts) Let P=[I0CI]P = \begin{bmatrix} I & 0 \\ C & I \end{bmatrix} and M=[A11A12A21A22]M = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} with conformable blocks. Compute the product PMPM in block form, and describe in words the "block row operation" that PP performs on MM.

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

    (a) If AA and BB are partitioned conformably for block multiplication, then the product ABAB can be computed as though the blocks were scalars.

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

    (c) A block-diagonal matrix diag⁡(A1,…,Ak)\operatorname{diag}(A_1, \ldots, A_k) is invertible if and only if every diagonal block AiA_i is invertible.

  6. 6. (3 pts) Using the column–row expansion of a product, write ABAB as a sum of outer products (column of AA times row of BB), and evaluate for

    A=[1234],B=[5678].A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \qquad B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}.

  7. 7. (2 pts) Explain why, for the block product [A11A12][B11B21]=A11B11+A12B21\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.

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

    M=[I0CA−1I][AB0S],M = \begin{bmatrix} I & 0 \\ CA^{-1} & I \end{bmatrix}\begin{bmatrix} A & B \\ 0 & S \end{bmatrix},

    and use it to explain why MM is invertible.