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Chapter 1: Problem 28

Let \(A\) be an \(m \times n\) matrix. Show that \(A^{T} A\) and \(A A^{T}\) are both symmetric.

Short Answer

Expert verified
We have shown that both \((A^{T}A)^{T} = A^{T}A\) and \((AA^{T})^{T} = AA^{T}\) using the properties of transpose and matrix multiplication. Therefore, the matrix products \(A^{T}A\) and \(AA^{T}\) are both symmetric.

Step by step solution

01

Determine \(A^{T}A\) and \((A^{T}A)^{T}\)

First, let's find \((A^{T}A)^{T}\) using the properties of transpose operations. Recall the property \((AB)^{T} = B^{T}A^{T}\), where A and B are matrices. Given \(A^{T}A\), we will find its transpose: \((A^{T}A)^{T} = (A)^{T}(A^{T})^{T}\)
02

Simplify \((A^{T}A)^{T}\)

Now, we simplify \((A)^{T}(A^{T})^{T}\), keeping in mind that \((A^{T})^{T} = A\): \((A)^{T}(A^{T})^{T} = (A)^{T}(A)\) Since \((A^{T}A)^{T}\) simplifies to \(A^{T}A\), we can say that \(A^{T}A\) is symmetric.
03

Determine \(AA^{T}\) and \((AA^{T})^{T}\)

Similar to step 1, let's find \((AA^{T})^{T}\) using the properties of transpose operations. Given \(AA^{T}\), we will find its transpose: \((AA^{T})^{T} = (A^{T})^{T}(A)^{T}\)
04

Simplify \((AA^{T})^{T}\)

Now, we simplify \((A^{T})^{T}(A)^{T}\), keeping in mind that \((A^{T})^{T} = A\): \((A^{T})^{T}(A)^{T} = (A)(A^{T})\) Since \((AA^{T})^{T}\) simplifies to \(AA^{T}\), we can say that \(AA^{T}\) is symmetric. #Conclusion#: We have shown that both \((A^{T}A)^{T} = A^{T}A\) and \((AA^{T})^{T} = AA^{T}\). Therefore, the matrix products \(A^{T}A\) and \(AA^{T}\) are both symmetric.

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Most popular questions from this chapter

Let \\[ A=\left(\begin{array}{ll} O & I \\ B & O \end{array}\right) \\] where all four submatrices are \(k \times k\). Determine \(A^{2}\) and \(A^{4}\)

If \(A\) and \(B\) are nonsingular matrices, then \((A B)^{T}\) is nonsingular and \\[ \left((A B)^{T}\right)^{-1}=\left(A^{-1}\right)^{T}\left(B^{-1}\right)^{T} \\]

Perform each of the following block multiplications: (a) \(\left[\begin{array}{ccc|c}1 & 1 & 1 & -1 \\ 2 & 1 & 2 & -1\end{array}\right)\left(\begin{array}{rrr}4 & -2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 3\end{array}\right)\) (b) \(\left(\begin{array}{rr}4 & -2 \\ 2 & 3 \\ \frac{1}{1} & 2\end{array}\right)\left(\begin{array}{rrr|r}1 & 1 & 1 & -1 \\ 2 & 1 & 2 & -1\end{array}\right)\) \((\mathbf{c})\left(\begin{array}{rr|rr}\frac{3}{5} & -\frac{4}{5} & 0 & 0 \\\ \frac{4}{5} & \frac{3}{5} & 0 & 0 \\ \hline 0 & 0 & 1 & 0\end{array}\right)\left(\begin{array}{rr|r}\frac{3}{5} & \frac{4}{5} & 0 \\\ -\frac{4}{5} & \frac{3}{5} & 0 \\ \hline 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right)\) (d) \(\left(\begin{array}{ccc|cc}0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0\end{array}\right)\left(\begin{array}{cc}1 & -1 \\ 2 & -2 \\ 3 & -3 \\ 4 & -4 \\ 5 & -5\end{array}\right)\)

Let \(A\) be a nonsingular matrix. Show that \(A^{-1}\) is also nonsingular and \(\left(A^{-1}\right)^{-1}=A\).

Let \(A\) be an \(m \times n\) matrix. Explain why the matrix multiplications \(A^{T} A\) and \(A A^{T}\) are possible.
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