## Linear Algebra and Its Applications, Exercise 3.3.7

Exercise 3.3.7. Given the two vectors $a_1 = (1, 0, 1)$ and $a_2 = (1, 1, -1)$ find the projection matrix $P$ that projects onto the subspace spanned by $a_1$ and $a_2$.

Answer: The subspace spanned by $a_1$ and $a_2$ is the column space $\mathcal{R}(A)$ where $A = \begin{bmatrix} 1&1 \\ 0&1 \\ 1&-1 \end{bmatrix}$

The projection matrix onto the subspace is then $P = A(A^TA)^{-1}A^T$. We have $A^TA = \begin{bmatrix} 1&0&1 \\ 1&1&-1 \end{bmatrix} \begin{bmatrix} 1&1 \\ 0&1 \\ 1&-1 \end{bmatrix} = \begin{bmatrix} 2&0 \\ 0&3 \end{bmatrix}$

Since $A^TA$ is a diagonal matrix we can compute its inverse by simply taking the reciprocals of the diagonal entries: $(A^TA)^{-1} = \begin{bmatrix} \frac{1}{2}&0 \\ 0&\frac{1}{3} \end{bmatrix}$

We then have $P = A(A^TA)^{-1}A^T = \begin{bmatrix} 1&1 \\ 0&1 \\ 1&-1 \end{bmatrix} \begin{bmatrix} \frac{1}{2}&0 \\ 0&\frac{1}{3} \end{bmatrix} \begin{bmatrix} 1&0&1 \\ 1&1&-1 \end{bmatrix}$ $= \begin{bmatrix} \frac{1}{2}&\frac{1}{3} \\ 0&\frac{1}{3} \\ \frac{1}{2}&-\frac{1}{3} \end{bmatrix} \begin{bmatrix} 1&0&1 \\ 1&1&-1 \end{bmatrix} = \begin{bmatrix} \frac{5}{6}&\frac{1}{3}&\frac{1}{6} \\ \frac{1}{3}&\frac{1}{3}&-\frac{1}{3} \\ \frac{1}{6}&-\frac{1}{3}&\frac{5}{6} \end{bmatrix}$

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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