## 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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