# Help with determinants and eigenvalues.

• Nov 16th 2010, 04:31 PM
pantsaregood
Help with determinants and eigenvalues.
Attachment 19739

For #1, I currently have that the two values are equal if A is either a matrix whose determinant is zero or a 1x1 matrix.

I don't really have any direction for the others. I see that the determinant of a square matrix whose columns add to zero is 0, but I'm not sure of why. Can it somehow be reduced to the zero matrix?

Edit: Okay, messed around with row reduction. Quickly found that all columns adding to zero forces at least one row to row-reduce to a zero row, forcing a determinant of zero.
• Nov 16th 2010, 04:37 PM
dwsmith
det(cA)=c^n*det(a) where n is the number of rows so as long as you have an odd number of rows and the scalar is -1, then det(-A)=-det(A)
• Nov 16th 2010, 04:45 PM
pantsaregood
I'm not following.

-det(A) = det(-A) and det(-A) = -1^n*det(A) don't immediately seem to be equivalent statements. In that case, a 2x2 matrix would have the property det(A) = det(-A).

Oh, okay. Nevermind. Odd number of rows.
• Nov 16th 2010, 05:34 PM
Drexel28
Quote:

Originally Posted by pantsaregood
I'm not following.

-det(A) = det(-A) and det(-A) = -1^n*det(A) don't immediately seem to be equivalent statements. In that case, a 2x2 matrix would have the property det(A) = det(-A).

Oh, okay. Nevermind. Odd number of rows.

2) If $\displaystyle c_1,\cdots,c_n$ are the columns it's asking for what $\displaystyle \det\left[\begin{array}{c|c|c}& & &\\ c_1 & \cdots & c_n\\ & & &\end{array}\right]$ given that $\displaystyle \displaystyle \sum_{j=1}^{n}c_j=\begin{bmatrix}0\\ \vdots\\ 0\end{bmatrix}$. Let me ask you this: what is the determinant of a matrix whose columns are linearly dependent equal? Moreover, how does that apply to this question?

3) This is pretty straightforward. I don't know how to give a decent hint that doesn't give it away. Merely note that $\displaystyle (AB)\bold{x}=A\left(B\bold{x}\right)$ and go from there.

4) If $\displaystyle \lambda$ is an eigenvalue for $\displaystyle A$ note that $\displaystyle \bold{0}=A^k\bold{x}=\lambda A^{k-1}\bold{x}=\cdots=\lambda^k\bold{x}$...so...