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Let $f:\mathbb{R}^3\rightarrow \mathbb{R}^3$ be an orthogonal projection on the plane $\pi:\{x+y+z=0\}$. Let $A$ be the image matrix of $f$ (w.r.t the standard basis). Determine col$(A),$ ker$(A),$ rank$(A)$ and nulldim$(A)$.

I usually give my own attempt when I ask questions here, but I'm really lost in these types of questions. Any push in the correct direction is welcome. How should one even begin to attack this problem?

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  • $\begingroup$ Hint: Choose $\mathcal{B} = \{(1,0,0)^T, (0,1,0)^T, (1,1,1)^T\}$ as your basis for $\mathbb{R}^3$, and compute the matrix $A$ of $f$ with respect to $\mathcal{B}$. Then find the change of basis matrix from $\mathcal{B}$ to the standard basis and conjugate $A$ by it. $\endgroup$ – André 3000 Oct 21 '17 at 13:43
  • $\begingroup$ You can think of this geometrically, and so you should already know the answer to each of these four things. For example, $f$ is clearly surjective, and so the column space of $A$ will be exactly the plane $\{x+y+z=0\}$. Since this is two dimensional, the rank of $A$ should be 2. So what kind of answer are you looking for? Are you trying to do the linear algebra to back this up? $\endgroup$ – Joppy Oct 21 '17 at 15:53
  • $\begingroup$ How is $f$ clearly surjective? What about kernel and nulldim? $\endgroup$ – Parseval Oct 21 '17 at 17:36

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