Physics Forums

## Diagonalization

1. The problem statement, all variables and given/known data

I think my teacher made a mistake in his homework answer. I need to verify this for practice. The answer I got is below. The answer the teacher has is in the pdf.

2. Relevant equations

3. The attempt at a solution

So there is two eigenvalues= 4 and 2
but the eigenvalue 2 has 2 eigenvectors [-1 1 0]T and [0 0 1]T but my teacher has only one [-1 1 0]T. That's why he says A is not diagonalizable. Do you think it's correct?
Attached Thumbnails

PhysOrg.com  science news on PhysOrg.com  >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug

I get the same result as your teacher for $$\lambda = 2$$  $$A= \begin{bmatrix} 3 & 1 & 0\\ 0 & 2 & 1\\ 1 & 1 & 3 \end{bmatrix}$$  so for lamda = 2, $$(A-2I)\vec{v}=\vec{0}$$ $$\begin{bmatrix} 1 & 1 & 0\\ 0 & 0 & 1\\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$$ I've personally always found it easier to not do row operations here and just jump straight in. From that you get $$v_3 = 0$$ and $$v_1 = -v_2$$ so therefore $$\vec{v} = \begin{bmatrix} 1\\ -1\\ 0 \end{bmatrix}$$  Since it is a 3x3 matrix, it needs 3 eigenvectors to be diagonalizable.  Hope this helps.

Hi, thanks for replying. Attached is how I got my vectors, do you think my steps are correct. 		 		  		 		 			  			 				 					Attached Thumbnails

Mentor

## Diagonalization

No. Why do you think (0, 0, 1) is an eigenvector? You seem to just pull that out of thin air.

You seem to have made a mistake in the step  $$(A-2I)\vec{v}=\vec{0}$$ Why is your second row 1 0 1 instead of 0 0 1 ? Have you said $$v_2= \delta$$ ? I'm not sure if I have read that correctly. If it is a delta, you can't say that unless the whole row equals zero. i.e. $$\begin{bmatrix} 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix}$$

Tags
algebra, diagonalization, eigenvalue, linear

Thread Tools

|                                      |                            |         |
|--------------------------------------|----------------------------|---------|
| Similar Threads for: Diagonalization |                            |         |
| Thread                               | Forum                      | Replies |
|                                      | Calculus & Beyond Homework | 0       |
|                                      | Calculus & Beyond Homework | 3       |
|                                      | Calculus & Beyond Homework | 5       |
|                                      | Quantum Physics            | 1       |
|                                      | Quantum Physics            | 2       |
{"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": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "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": 0.9221944808959961, "perplexity_flag": "middle"}