Prove that for a real matrix $A$, $\ker (A) = \ker (A^TA)$
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linear algebra - Prove that $\ker (AB) = \ker (A) + \ker (B ...
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$\ker (A^TA) = \ker (A)$ - Mathematics Stack Exchange
Sep 9, 2018 · I am sorry i really had no clue which title to choose. I thought about that matrix multiplication and since it does not change the result it is idempotent? Please suggest a better …
Can $RanA=KerA^T$ for a real matrix $A$? And for complex $A$?
Jan 27, 2021 · It does address complex matrices in the comments as well. It is clear that this can happen over the complex numbers anyway.
Prove that $\operatorname {im}\left (A^ {\top}\right) = \ker (A ...
I need help with showing that $\ker\left (A\right)^ {\perp}\subseteq Im\left (A^ {T}\right)$, I couldn't figure it out.
How to find $ker (A)$ - Mathematics Stack Exchange
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If Ker (A)=$\ {\vec {0}\}$ and Ker (B)=$\ {\vec {0}\}$ Ker (AB)=?
Thank you Arturo (and everyone else). I managed to work out this solution after completing the assigned readings actually, it makes sense and was pretty obvious. Could you please comment …
linear algebra - proof of KerA = ImB implies ImA^T = KerB^T ...
Sep 3, 2019 · proof of KerA = ImB implies ImA^T = KerB^T Ask Question Asked 6 years ago Modified 6 years ago
How to find basis of ker (A)? - Mathematics Stack Exchange
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How to find a basis of an image of a linear transformation?
It is $$ kerA = (1,1,1) $$ But how can I find the basis of the image? What I have found so far is that I need to complement a basis of a kernel up to a basis of an original space. But I do not have an …
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