## Linear Algebra and Its Applications, Exercise 3.3.10

Exercise 3.3.10. Given mutually orthogonal vectors $a_1$, $a_2$, and $b$ and the matrix $A$ with columns $a_1$ and $a_2$, what are $A^TA$ and $A^Tb$? What is the projection of $b$ onto the plane formed by $a_1$ and $a_2$? $A^TA = \begin{bmatrix} a_1^T \\ a_2^T \end{bmatrix} \begin{bmatrix} a_1&a_2 \end{bmatrix} = \begin{bmatrix} a_1^Ta_1&a_1^Ta_2 \\ a_2^Ta_1&a_2^Ta_2 \end{bmatrix} = \begin{bmatrix} \|a_1\|^2&0 \\ 0&\|a_2\|^2 \end{bmatrix}$

where the zero entries are the result of $a_1$ and $a_2$ being orthogonal.

Similarly we have $A^Tb = \begin{bmatrix} a_1^T \\ a_2^T \end{bmatrix} b = \begin{bmatrix} a_1^Tb \\ a_2^Tb \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0$

where the zero entries are the result of $a_1$ and $a_2$ being orthogonal to $b$.

Since $b$ is orthogonal to both $a_1$ and $a_2$ it is orthogonal to any linear combination of $a_1$ and $a_2$ and therefore is orthogonal to the plane spanned by $a_1$ and $a_2$. The projection of $b$ onto that plane is therefore the zero vector.

This also follows from the formula for the projection matrix $P$ corresponding to the matrix $A$. The projection $p$ of $b$ onto the column space of $A$ (the space spanned by $a_1$ and $a_2$) is $p = Pb = A(A^TA)^{-1}A^Tb = A(A^TA)^{-1} \cdot 0 = 0$

since $A^Tb = 0$ as discussed above.

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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