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Today we were discussing how for an nxn orthogonal projection matrix from $\mathbb{R^{n}}$ onto a subspace W, Ker($A$)=$(Im$A$)^{\perp}$=$W^{\perp}$ and that Ker($A^{T}$) is also $W^{\perp}$. This prompted the question of what conditions are necessary for a matrix so that the kernels of it and its transpose are equal. It looks like it always works when it's an orthogonal projection, but we struggled to find an example of a square matrix for which the aforementioned identities would not hold, and consequently for which Ker($A$)$\neq$Ker($A^{T}$). It looks like we need to find a transformation whose Kernel would not consist only of things perpendicular to its image, but we were wondering if that was possible or if our reasoning was correct at all. Could you please clarify this issue (I know it's convoluted) and try to point out where we were right and wrong, and when do those identities hold?

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By "the projection", you clearly mean the orthogonal projection... You might want to put it explicitly, though. – Arturo Magidin Mar 11 '11 at 5:28
@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). – LinAlgStudent Mar 11 '11 at 5:31
@LiinAlgStudent: You are definitely talking about the orthogonal projection. In general, given a vector space $\mathbf{V}$, if you write $\mathbf{V}=\mathbf{W}\oplus\mathbf{U}$, you get a projection onto $\mathbf{W}$ (which depends on $U$): given any $\mathbf{v}\in\mathbf{V}$ there is a unique $\mathbf{w}\in\mathbf{W}$ and a unique $\mathbf{u}\in\mathbf{U}$ such that $\mathbf{v}=\mathbf{w}+\mathbf{u}$. Map $\mathbf{v}$ to the corresponding $\mathbf{w}$. When $\mathbf{U}=\mathbf{W}^{\perp}$, you get the orthogonal projection onto $\mathbf{W}$. – Arturo Magidin Mar 11 '11 at 5:45
up vote 2 down vote accepted

Things always work nicely for projections, because unlike other matrices we have $\text{Range}\oplus\text{kernel}=\text{whole space}$ Let $V=\mathbb{R}^2$ Consider $$A=\left[\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right]$$

Then $$\text{ker}(A)=\left\{ \left[\begin{array}{c} a\\ 0\end{array}\right]\ :\ a\in\mathbb{R}\right\},$$ but $$A^{t}=\left[\begin{array}{cc} 0 & 0\\ 1 & 0\end{array}\right]$$ so that $$\text{ker}(A^{t})=\left\{ \left[\begin{array}{c} 0\\ a\end{array}\right]\ :\ a\in\mathbb{R}\right\}.$$ These are definitely different. (In contrast to the previous example, their direct sum is the whole space $V$)

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Ok, so right off the bat, Ker A is not equal to the perp of the image, so there's an example of a non-projection matrix that suppresses the dimension (for some reason didn't see it). Thanks a lot. – LinAlgStudent Mar 11 '11 at 5:33

$A$ and $A^{T}$ are similar. So the answer should be the same. You can try standard Jordan decomposition and see how the Jordan block works. A few cases in $2\times 2$ would be very helpful.

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So Ker(A) is always equal to Ker(A) transpose, do I understand you correctly? Not sure what Jordan block is, we are just beginners. – LinAlgStudent Mar 11 '11 at 5:20
yes - I am also a beginner. I would be glad to know if I were wrong and someone can pointed me out. For Jordan decomposition you can see here: – Kerry Mar 11 '11 at 5:22
I will look into that, but it seems like a bold claim -- I think that would be a very important property for any square matrix if its kernel was equal to the kernel of its transpose. – LinAlgStudent Mar 11 '11 at 5:25
@LinAlgStudent: It is not true in general. Take $$A=\left(\begin{array}{cc}0&1\\0&0\end{array}\right), \quad A^T=\left(\begin{array}{cc}0&0\\1&0\end{array}\right).$$ The kernel of $A$ is $\{(x,0)\mid x\in\mathbb{R}\}$, the kernel of $A^T$ is $\{(0,y)\mid y\in\mathbb{R}\}$. – Arturo Magidin Mar 11 '11 at 5:41

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