These are True/False concept questions from Howard Anton's Linear Algebra, 11th edition, contained in the first 200 pages of the book. I did some 60 of them and I got these wrong and I'm not sure what I'm doing wrong:

I will continually delete questions as I get answers

5. If A and B are $nXn$ matrices such that $A + B$ is symmetric, then A and B are symmetric.

6. If A is a 4x4 matrix and B is obtained from A by swap of the first two lines, followed by the swap of the last two lines, then $det(B) = det(A)$. The official answer is False.

7. If A is a 3x3 matrix and B is a matrix obtained from A by multiplying the first line by 4, then the last by 3/4, then $det(B) = 3det(A)$. The official answer is False.

8. If $A²$ is a symmetric matrix, then A is symmetric.

My reasoning

5.I don't know how I would prove that. If I start creating two matrices A and B, both symmetric, then I can show that A + B is symmetric, but the problem is showing the other way around, that is, you need to come up with one symmetric matrix and decompose it into two symmetric matrices and I don't know how to do that.

6.Yes, swap two lines and the sign of the determinant changes. Swap two lines again and the sign of the determinant comes back just as in the original matrix.

7.Yes, in calculating the determinant you could factor out a 4, then factor out a 3/4. Then $4*(3/4) = 3$.

8.I can start with a symmetric matrix and show that if you square it, A² is symmetric, but then again, what I need is the other way around, and I don't know how to proceed here.

  • $\begingroup$ 1. is true but I can't find any logic in your reasoning. This claim is true because in a square matrix, upon taking transpose the main diagonal remmins unchanged. $\endgroup$ – DonAntonio Apr 9 '18 at 19:38
  • $\begingroup$ In 1, $tr$ means "trace" not "transpose". $\endgroup$ – saulspatz Apr 9 '18 at 19:42
  • $\begingroup$ Take matrix A. Transpose it. Transpose it again. What you get is the original A, right? That's what i thought but the correct answer is False. $\endgroup$ – Victor S. Apr 9 '18 at 19:43
  • $\begingroup$ Trace... wow. I thought it was another notation for "transpose". Thank you. $\endgroup$ – Victor S. Apr 9 '18 at 19:44
  • $\begingroup$ That first item, the multiplier was the product of the n first numbers, not the sum. $\endgroup$ – Victor S. Apr 9 '18 at 22:27

For (2): just substitute:

$$A(Sx)=Sb\iff S^{-1}ASx=B$$

For (4):

$$Ax=4x\iff (A-4I)x=0\;\;\text{has unique solution}\iff A-4I\;\;\text{invertible}$$

The only solution being $\;\vec x=0\;$ , of course.

For (5)

$$\begin{pmatrix}0&1\\0&0\end{pmatrix}+\begin{pmatrix}0&0\\1&0\end{pmatrix}\;\;\text{is symmetric...}$$

For (8):

$$\begin{pmatrix}0&1\\0&0\end{pmatrix}^2\;\;\;\text{is symmetric}$$

  • $\begingroup$ Unless I’m confused, your hint for (5) doesn’t seem to work. The two matrices are symmetric... $\endgroup$ – Theoretical Economist Apr 9 '18 at 19:45
  • $\begingroup$ @TheoreticalEconomist Of course, thanks. I got confused between questions...:). Editing $\endgroup$ – DonAntonio Apr 9 '18 at 19:45
  • $\begingroup$ In 3 i realized I could Multiply by S by the left on both sides, arriving at ASx = Sb. So you just plug y = Sx as in "y" is a variable? wow... thanks. $\endgroup$ – Victor S. Apr 9 '18 at 19:56
  • $\begingroup$ In 5, yes, the sum is symmetric but each individual matrix is not. $\endgroup$ – Victor S. Apr 9 '18 at 20:17
  • $\begingroup$ Again, A² is symmetric but A is not symmetric. $\endgroup$ – Victor S. Apr 9 '18 at 20:27

You have missed these questions for a good reason.

You need to refresh your memory on some definitions.

For example, $$tr(A^t) = tr(A)$$ here the tr(A) stands for the trace of A not the transpose of A.

The permutation of rows does not mean the row swap. It means to permute the elements of one row as a one-to-one function of the other.

  • $\begingroup$ I agree with the OP's interpretation of question 2. Compare question 6. $\endgroup$ – saulspatz Apr 9 '18 at 19:54
  • $\begingroup$ Yes, I just learned that tr means "trace". I thought it was another notation for transpose. Thank you. $\endgroup$ – Victor S. Apr 9 '18 at 19:58
  • $\begingroup$ By the way the "permutation" means swap really, any confusion may be because I'm translating from portuguese. $\endgroup$ – Victor S. Apr 9 '18 at 20:10

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