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

• Dec 6th 2010, 02:33 PM
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!
• Dec 6th 2010, 02:47 PM
dwsmith
If the matrix isn't square, then you have to find the Ker and Im every time without short cuts when transposed.
• Dec 6th 2010, 02:48 PM
phoo :(
and if it is square?
• Dec 6th 2010, 02:51 PM
dwsmith
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.
• Dec 6th 2010, 03:17 PM
Drexel28
Quote:

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 $\ker(A^{\top})=\left(\text{im}(A)\right)^{\perp}$.
• Dec 6th 2010, 04:02 PM
Quote:

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 :(
• Dec 8th 2010, 07:25 PM
can someone please help how do you find the orthogonal projections onto the kernel and image of A transposed?
• Dec 9th 2010, 11:26 AM
nobody?
• Dec 9th 2010, 12:18 PM
dwsmith
Post your 4x4 matrix A with what you have done.
• Dec 9th 2010, 02:24 PM
[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
• Dec 9th 2010, 02:30 PM
dwsmith
Yes, use the GS process.
• Dec 9th 2010, 02:40 PM
dwsmith
$\displaystyle
\begin{bmatrix}
1 & 1 & 0 & 1\\
1 & 0 & 3 & -1\\
1 & 0 & 1 & -1\\
1 & 2 & 0 & 3
\end{bmatrix}\rightarrow \ \mbox{rref}=
\begin{bmatrix}
1 & 0 & 0 & -1\\
0 & 1 & 0 & 2\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}\rightarrow \ x_1=x_4, \ x_2=-2x_4, \ x_3=0, \ x_4$

Basis for the Kernel: $\displaystyle \ \ \ x_4\begin{bmatrix}
1\\
-2\\
0 \\
1
\end{bmatrix}$
• Dec 9th 2010, 03:17 PM