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

**giygaskeptpraying** · December 6th 2010, 02:33 PM

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!

**dwsmith** · December 6th 2010, 02:47 PM

If the matrix isn't square, then you have to find the Ker and Im every time without short cuts when transposed.

**giygaskeptpraying** · December 6th 2010, 02:48 PM

phoo

and if it is square?

**dwsmith** · December 6th 2010, 02:51 PM

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.

**Drexel28** · December 6th 2010, 03:17 PM

Originally Posted by giygaskeptpraying

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 $\ker(A^{\top})=\left(\text{im}(A)\right)^{\perp}$ .

**giygaskeptpraying** · December 6th 2010, 04:02 PM

Originally Posted by Drexel28

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

**giygaskeptpraying** · December 8th 2010, 07:25 PM

can someone please help how do you find the orthogonal projections onto the kernel and image of A transposed?

**giygaskeptpraying** · December 9th 2010, 11:26 AM

nobody?

**dwsmith** · December 9th 2010, 12:18 PM

Post your 4x4 matrix A with what you have done.

**giygaskeptpraying** · December 9th 2010, 02:24 PM

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

**dwsmith** · December 9th 2010, 02:30 PM

Yes, use the GS process.

**dwsmith** · December 9th 2010, 02:40 PM

$\displaystyle \begin{bmatrix} 1 & 1 & 0 & 1\\ 1 & 0 & 3 & -1\\ 1 & 0 & 1 & -1\\ 1 & 2 & 0 & 3 \end{bmatrix}\rightarrow \ \mbox{rref}= \begin{bmatrix} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\rightarrow \ x_1=x_4, \ x_2=-2x_4, \ x_3=0, \ x_4$

Basis for the Kernel: $\displaystyle \ \ \ x_4\begin{bmatrix} 1\\ -2\\ 0 \\ 1 \end{bmatrix}$

**giygaskeptpraying** · December 9th 2010, 03:17 PM

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?

**dwsmith** · December 9th 2010, 03:19 PM

Not sure but take that and see what you get.

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