# Help with determinants and eigenvalues.

Printable View

• November 16th 2010, 04:31 PM
pantsaregood
1 Attachment(s)
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.
• November 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)
• November 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.
• November 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 $c_1,\cdots,c_n$ are the columns it's asking for what $\det\left[\begin{array}{c|c|c}& & &\\ c_1 & \cdots & c_n\\ & & &\end{array}\right]$ given that $\displaystyle \sum_{j=1}^{n}c_j=\begin{bmatrix}0\\ \vdots\\ 0\end{bmatrix}$. 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 $(AB)\bold{x}=A\left(B\bold{x}\right)$ and go from there.

4) If $\lambda$ is an eigenvalue for $A$ note that $\bold{0}=A^k\bold{x}=\lambda A^{k-1}\bold{x}=\cdots=\lambda^k\bold{x}$...so...
All times are GMT -8. The time now is 10:53 PM.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9178259372711182, "perplexity_flag": "middle"}