## Linear Algebra and Its Applications, Exercise 3.2.10

Exercise 3.2.10. Discuss whether the projection matrix $P$ from exercise 3.2.9 is invertible or not. Explain your answer. $P = aa^T/a^Ta = \frac{1}{a_1a_1 + a_2a_2 + \dots + a_na_n} \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix}\begin{bmatrix} a_1&a_2&\dots&a_n \end{bmatrix}$ $= \frac{1}{a_1a_1 + a_2a_2 + \dots + a_na_n} \begin{bmatrix} a_1 \cdot a_1&a_1 \cdot a_2&\cdots&a_1 \cdot a_n \\ a_2 \cdot a_1&a_2 \cdot a_2&\cdots&a_2 \cdot a_n \\ \vdots&\vdots&\ddots&\vdots \\ a_n \cdot a_1&a_n \cdot a_2&\cdots&a_n \cdot a_n \end{bmatrix}$

Note that each column $i$ of the resulting matrix is equal to the original vector $a = \left(a_1, a_2, \dots, a_n \right)$ multiplied by a scalar factor equal to $a_i/a^Ta$. The columns are therefore linearly dependent with each of them expressible as the first column times the factor $a_i/a_1$, and thus the rank of the projection matrix $P$ is $r = 1$. Since the rank $r < n$ the matrix $P$ is singular and not invertible.

Also note that this is consistent with the geometry of the situation: Since $P$ projects vectors from a higher-dimensional vector space onto a line, many points in the space will end up being mapped into a single point on a line. There is thus no way to reverse the mapping and map a point on the line onto a single unique vector in the higher-dimensional space.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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