Timeline for Transforming a non-invertible matrix into an invertible matrix
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 6, 2016 at 23:19 | answer | added | Nawaf Bou-Rabee | timeline score: 3 | |
| Oct 5, 2016 at 1:59 | comment | added | Nawaf Bou-Rabee | @IlyaBogdanov: so this projector obliquely projects onto the kernel of A along the image of A. That makes perfect sense now. Thanks! | |
| Oct 4, 2016 at 14:58 | comment | added | Ilya Bogdanov | @NawafBou-Rabee: the projector is $\left[\begin{matrix}0& -1\\0& 1\end{matrix}\right]$: it maps $(1,0)^T$ to $0$, but $(1,-1)^T$ to itself. The result is thus $\left[\begin{matrix}1& 0\\0& 1\end{matrix}\right]$. | |
| Oct 4, 2016 at 14:08 | comment | added | Nawaf Bou-Rabee | @IlyaBogdanov can you specify your B for the 2x2 example from my previous comment? | |
| S Oct 3, 2016 at 12:44 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor edits
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| Oct 3, 2016 at 12:32 | review | Suggested edits | |||
| S Oct 3, 2016 at 12:44 | |||||
| Oct 2, 2016 at 8:13 | comment | added | Ilya Bogdanov | @Nawai: My projector is not orthogonal: it is along $\mathop{\rm Im} A$, so all the vectors in the image are mapped to zero. | |
| Oct 1, 2016 at 21:14 | review | Close votes | |||
| Oct 1, 2016 at 22:24 | |||||
| Oct 1, 2016 at 20:13 | comment | added | Nawaf Bou-Rabee | Ilya: the matrix $A$ is not necessarily orthogonally diagonalizable. Consider, e.g., $A = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$, which is perfectly diagonalizable with null space in the anti-diagonal direction; Yet, you can check that $B = A + \text{Proj}_{N(A)}$ no longer has the eigenvector $(1,0)^T$ of $A$, since $(1,0)^T$ has a component in the anti-diagonal direction. | |
| Oct 1, 2016 at 19:35 | comment | added | Julian Rosen | Let $f(T)$ be the characteristic polynomial of $A$ with all the factors of $T$ divided out, and take $B=A+f(A)$. | |
| Oct 1, 2016 at 18:58 | comment | added | Ilya Bogdanov | Add the projector onto $\mathop{\rm Ker} A$ along $\mathop{\rm Im} A$... | |
| Oct 1, 2016 at 17:55 | history | edited | Dudi Frid | CC BY-SA 3.0 |
added 109 characters in body
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| Oct 1, 2016 at 17:29 | review | First posts | |||
| Oct 1, 2016 at 17:30 | |||||
| Oct 1, 2016 at 17:25 | history | asked | Dudi Frid | CC BY-SA 3.0 |