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- 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?
- linear algebra - Prove that $\ker (AB) = \ker (A) + \ker (B . . .
It is not right: take $A=B$, then $\ker (AB)=\ker (A^2) = V$, but $ker (A) + ker (B) = ker (A)$
- $\ker (A^TA) = \ker (A)$ - Mathematics Stack Exchange
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
- Can $RanA=KerA^T$ for a real matrix $A$? And for complex $A$?
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
- 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 I know that Ker (A)=Ker (rref (A)) for any matrix A, I am not sure if I can say that Ker (rref (A) * rref (B))=Ker (AB) Is this statement true?" just out of my curiosity?
- 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 with my keyboard sorry! We have ker (A)= {x∈V:A⋅x=0} This means if a vector x when applied to our system of equations (Matrix) are takin to the Zero Vector This question is
- linear algebra - A confusion about Ker ($A$) and Ker ($A^ {T . . .
@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 projection, though I could be wrong (maybe we are just assumed not to do any other projections at our level, or maybe it was assumed it was a perpendicular projection, which I guess is the same thing)
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