## 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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