## Linear Algebra and Its Applications, Exercise 3.3.14

Exercise 3.3.14. Find the projection matrix $P$ onto the plane spanned by the vectors $(1, 1, 0)$ and $(1, 1, 1)$. Find a nonzero vector $b$ that $P$ projects to zero.

Answer: The plane in question is the column space of the matrix $A = \begin{bmatrix} 1&1 \\ 1&1 \\ 0&1 \end{bmatrix}$

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

so that its inverse is $(A^TA)^{-1} = \frac{1}{2 \cdot 3 - 2 \cdot 2} \begin{bmatrix} 3&-2 \\ -2&2 \end{bmatrix}$ $= \frac{1}{2} \begin{bmatrix} 3&-2 \\ -2&2 \end{bmatrix} = \begin{bmatrix} \frac{3}{2}&-1 \\ -1&1 \end{bmatrix}$

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

This completes the first part of the exercise. For the second part, we need to find a nonzero vector $b$ projected into zero by $P$. This will be true if $b$ is in the left nullspace of $A$, since $b$ will then be orthogonal to the plane corresponding to the column space of $A$.

We therefore look for a vector in the left nullspace $\mathcal{N}(A^T)$ by solving the system $A^Ty = 0$: $\begin{bmatrix} 1&1&0 \\ 1&1&1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = 0$

We perform Gaussian elimination by multiplying the first row by 1 and subtracting it from the second row: $\begin{bmatrix} 1&1&0 \\ 1&1&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&0 \\ 0&0&1 \end{bmatrix}$

The resulting matrix is in echelon form, with pivots in columns 1 and 3. Thus $y_1$ and $y_3$ are basic variables, and $y_2$ is a free variable. Setting $y_2 = 1$ we have $y_3 = 0$ (from the second row) and $y_1 = -1$ (from the first row). The vector $b = (-1, 1, 0)$ is therefore a solution to $A^Ty = 0$ and is a basis for the left nullspace of $A$.

Since $b$ is in $\mathcal{N}(A^T)$ it is orthogonal to $\mathcal{R}(A)$ and is therefore projected by $P$ into zero: $Pb = \begin{bmatrix} \frac{1}{2}&\frac{1}{2}&0 \\ \frac{1}{2}&\frac{1}{2}&0 \\ 0&0&1& \end{bmatrix} \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

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, Fifth Edition and the accompanying free online course, and Dr Strang’s other books .

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