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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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    @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 $\mathbf{W}$.2011-03-11

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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 $n$ot equal to the perp of the image, so there's a$n$ example of a $n$o$n$-projection matrix that suppresses the dimension (for some reason didn't see it). Thanks a lot.2011-03-11
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$A$ 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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    @user7887: Yes, they $a$re "isomorphic", but th$a$t tells you precious little: it just tells you they have the same dime$n$sio$n$.2011-03-11
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I think the condition you will want to check for the kernel of a matrix to be equal to the kernel of the transpose of the matrix is when the matrix is symmetric.