## Linear Algebra and Its Applications, Exercise 1.16.15

Exercise 1.6.15. For any square matrix B and matrices A and K where

$A = B + B^T, \quad K = B - B^T$

prove that A is symmetric and K is skew-symmetric, i.e.,

$K^T = -K$

For the case where

$B = \begin{bmatrix} 1&3 \\ 1&1 \end{bmatrix}$

compute A and K and show that B can be expressed as the sum of a symmetric matrix and a skew-symmetric matrix.

Answer: Let $C = B^T$. For all elements of A we then have

$a_{ij} = b_{ij} + c_{ij} = c_{ji} + b_{ji} = b_{ji} + c_{ji} = a_{ji}$

so that

$A = A^T$

and A is a symmetric matrix.

For all elements of K we have

$k_{ij} = b_{ij} - c_{ij} = c_{ji} - b_{ji} = -(b_{ji} - c_{ji}) = -k_{ji}$

so that

$K^T = -K$

and K is a skew-symmetric matrix.

We then have

$B = \begin{bmatrix} 1&3 \\ 1&1 \end{bmatrix} \quad B^T = \begin{bmatrix} 1&1 \\ 3&1 \end{bmatrix}$

so that

$A = B + B^T = \begin{bmatrix} 1&3 \\ 1&1 \end{bmatrix} + \begin{bmatrix} 1&1 \\ 3&1 \end{bmatrix} = \begin{bmatrix} 2&4 \\ 4&2 \end{bmatrix}$

and

$K = B - B^T = \begin{bmatrix} 1&3 \\ 1&1 \end{bmatrix} - \begin{bmatrix} 1&1 \\ 3&1 \end{bmatrix} = \begin{bmatrix} 0&2 \\ -2&0 \end{bmatrix}$

Note that

$A + K = (B + B^T) + (B - B^T) = B + B + B^T - B^T = 2B$

so that

$B = \frac{1}{2}(A + K)$

In our example we then have

$\begin{bmatrix} 1&3 \\ 1&1 \end{bmatrix} = \frac{1}{2}(\begin{bmatrix} 2&4 \\ 4&2 \end{bmatrix} + \begin{bmatrix} 0&2 \\ -2&0 \end{bmatrix})$

$= \frac{1}{2} \begin{bmatrix} 2&4 \\ 4&2 \end{bmatrix} + \frac{1}{2} \begin{bmatrix} 0&2 \\ -2&0 \end{bmatrix} = \begin{bmatrix} 1&2 \\ 2&1 \end{bmatrix} + \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix}$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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