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i) Consider a matrix $A\in R^{{5\times2}}$ and a matrix $B\in R^{{5\times3}}$. It is given that $\operatorname{rank}(A)=2$ and $\operatorname{rank}(B)=3$. It is also given $A^TB$ is the $2\times3$ zero matrix, that is, the columns of $A$ are orthogonal to the columns of $B$. Determine the eigenvalues and corresponding eigenvectors of the matrix:

$$\Pi = A(A^TA)^{{-1}}A^T$$

ii) Assume again that $A\in R^{{5\times2}}$ has linearly independent columns and that $C$ is a $2\times2$ invertible matrix. Explain why the orthogonal projection onto the range of $AC$ is also the same.

From properties of orthogonal projection matrices, I am aware that the $\lambda=1,0$, however I am unsure how to determine eigenvectors. Thanks in advance!

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$ Commented Mar 16, 2023 at 21:09

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i) $\Pi$ has two different eigenvalues $\lambda = 0,1$. By calculating $\Pi A$ and $\Pi B$, you can find out that the columns of $A$ are eigenvectors corresponding to $\lambda = 1$, and the columns of $B$ to $\lambda = 0$.

Note: to determine the eigvenvectors here means to determine how the eigenvectors depend on $A$ and $B$. E.g.,

  1. take $U \in \mathbb{R}^{5 \times 5}$ be s.t. $U^T U$ is the identity matrix, and assume $U$ has columns $\vec{u}_1,\ldots,\vec{u}_5$;
  2. if $A$ consists of $\vec{u}_1,\vec{u}_2,\vec{u}_3$ and $B$ the remaining columns then the eigenvectors are $\vec{u}_1,\vec{u}_2,\vec{u}_3$ and $\vec{u}_4,\vec{u}_5$ (or their linear combinations).

So, there is no way to find 'concrete' eigenvectors.

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  • $\begingroup$ Hello, yes thank you! Unfortunately A and B are not defined in this problem other than by the properties mentioned in the problem statement. I am familiar with the traditional methods of calculating eigenvectors from corresponding eigen values, but unsure of how to do so in this particular problem. Thanks again! $\endgroup$ Commented Mar 17, 2023 at 15:55
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    $\begingroup$ Hello, thank you for your update! I see about not being able to find 'concrete' eigenvectors. I think in general suffices so your answer is very helpful! Appreciate it! $\endgroup$ Commented Mar 18, 2023 at 15:17

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