Linear Algebra and Its Applications, Exercise 1.6.14

Exercise 1.6.14. For any m x n matrix A, prove that AA^T and A^TA are symmetric matrices. Provide an example where these matrices are not equal.

Answer: Per Equation 1M(i) on page 47 we have

(AB)^T = B^TA^T

Substituting A^T for B we have

(AA^T)^T = (A^T)^TA^T = AA^T

Since AA^T is equal to its own transpose it is a symmetric matrix.

Similarly we have

(A^TA)^T = A^T(A^T)^T = A^TA

Since A^TA is equal to its own transpose it too is a symmetric matrix.

Next, we pick an example 2 x 2 matrix

A = \begin{bmatrix} 1&2 \\ 1&3 \end{bmatrix}

We then have

AA^T = \begin{bmatrix} 1&2 \\ 1&3 \end{bmatrix} \begin{bmatrix} 1&1 \\ 2&3 \end{bmatrix} = \begin{bmatrix} 5&7 \\ 7&10 \end{bmatrix}

and

A^TA = \begin{bmatrix} 1&1 \\ 2&3 \end{bmatrix} \begin{bmatrix} 1&2 \\ 1&3 \end{bmatrix} = \begin{bmatrix} 2&5 \\ 5&13 \end{bmatrix}

So in general AA^T does not equal A^TA.

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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