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Math Help - simple way to find basis for Ker(A-transpose) and Im(A-transpose)

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    simple way to find basis for Ker(A-transpose) and Im(A-transpose)

    I'm pretty sure there have to be some easier ways of finding bases for those two spaces other than taking the transpose of a given matrix A and row reducing it again (after row reducing the original A to find Im(A) and Ker(A)). Does anyone know? help would be appreciated. thanks!
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    If the matrix isn't square, then you have to find the Ker and Im every time without short cuts when transposed.
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    phoo
    and if it is square?
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  4. #4
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    If it is square, the same vectors should span then. I am almost positive on this but I could be wrong. I can't think of any examples where it doesn't work yet.
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  5. #5
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    Quote Originally Posted by giygaskeptpraying View Post
    I'm pretty sure there have to be some easier ways of finding bases for those two spaces other than taking the transpose of a given matrix A and row reducing it again (after row reducing the original A to find Im(A) and Ker(A)). Does anyone know? help would be appreciated. thanks!
    If you're talking about A:\mathbb{R}^n\to\mathbb{R}^m then \ker(A)=\left(\text{im}(A^{\top})\right)^{\perp} (with the usual inner product) and thus consequently \ker(A^{\top})=\left(\text{im}(A)\right)^{\perp}.
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    Quote Originally Posted by Drexel28 View Post
    If you're talking about A:\mathbb{R}^n\to\mathbb{R}^m then \ker(A)=\left(\text{im}(A^{\top})\right)^{\perp} (with the usual inner product) and thus consequently \ker(A^{\top})=\left(\text{im}(A)\right)^{\perp}.
    hmm for my problem I was given a 4x4 matrix A and had to find the orthogonal projections onto Ker(A transpose) and Im(A transpose)
    so I thought I'd have to find bases for them to calculate the projection
    I'm not sure how to use the equality you brought up to do it
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    can someone please help how do you find the orthogonal projections onto the kernel and image of A transposed?
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    nobody?
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    Post your 4x4 matrix A with what you have done.
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    the matrix A was this:
    [1 1 0 1
    1 0 3 -1
    1 0 1 -1
    1 2 0 3]
    I found a basis for Im(A) to be the first 3 columns of A and a basis for Ker(A) to be the vector [-3 0 2 3].
    I now have to find the orthogonal projections onto the kernal and image of A-transposed

    another question, not related to my original one:
    it says to find an orthonormal basis for Im(A)
    would you just use the gram-schmidt process? I'm given a hint that points out that two vectors of the basis I found for Im(A) are already orthogonal but I don't know what to do with it
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  11. #11
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    Yes, use the GS process.
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  12. #12
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    \displaystyle<br />
\begin{bmatrix}<br />
1 & 1 & 0 & 1\\ <br />
1 & 0 & 3 & -1\\ <br />
1 & 0 & 1 & -1\\ <br />
1 & 2 & 0 & 3<br />
\end{bmatrix}\rightarrow \ \mbox{rref}=<br />
\begin{bmatrix}<br />
1 & 0 & 0 & -1\\ <br />
0 & 1 & 0 & 2\\ <br />
0 & 0 & 1 & 0\\ <br />
0 & 0 & 0 & 0<br />
\end{bmatrix}\rightarrow \ x_1=x_4, \ x_2=-2x_4, \ x_3=0, \ x_4

    Basis for the Kernel: \displaystyle \ \ \ x_4\begin{bmatrix}<br />
1\\ <br />
 -2\\ <br />
0 \\ <br />
1  <br />
\end{bmatrix}
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  13. #13
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    is the only way to find the orthogonal projections onto the kernel and image of A transpose to take the A transpose and row reduce it all over again?
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  14. #14
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    Not sure but take that and see what you get.
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