Prove that for a real matrix $A$, $\ker (A) = \ker (A^TA)$
Is it possible to solve this supposing that inner product spaces haven't been covered in the class yet?
How to find $ker (A)$ - Mathematics Stack Exchange
So before I answer this we have to be clear with what objects we are working with here. Also, this is my first answer and I cant figure out how to actually insert any kind of equations, besides what I can type …
$\ker (A^TA) = \ker (A)$ - 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 one.
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 on "Also, while …
Prove that if $\ker (A) = \ker (A^2 )$, then $ \ker (A^k ) = \ker (A ...
Feb 6, 2015 · Let $b$ be a vector such that $Ab =0$, and $b$ is that kernel. Let's call that kernel $A$. Then $b$ is also the same as $\\ker(A^2)$. Any hints would be appreciated ...
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.
linear algebra - A confusion about Ker ($A$) and Ker ($A^ {T ...
May 9, 2024 · @Arturo: We actually got this example from the book, where it used projection on W to prove that dimensions of W + W perp are equal to n, but I don't think it mentioned orthogonal …
linear algebra - $\dim (\ker (ab)) = \dim (\ker (a)) + \dim (\ker (b ...
Feb 7, 2019 · Can somebody help me please so solve a self-study problem $V$ is a vector space over $F$ and $a,b$ are in $L (V)$. Suppose $\ker (a)$ and $\ker (b)$ are finite ...
linear algebra - What does $\ker (A) \cap \operatorname {im} (A)$ say ...
Oct 11, 2023 · I'm trying to investigate what $\\ker(A) \\cap \\operatorname{im}(A)$ implies for $\\operatorname{rank}(A)$ and $\\operatorname{null}(A)$. Specifically, if we have ...
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.