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The matrix is as follows:

$$ \begin{bmatrix} 3/5 & 4/5 & 3/5 \\ -4/5 & 3/5 & 0 \\ 0 & 0 & 4/5 \\ \end{bmatrix} $$

I get that in order for a matrix (call this matrix A) to be orthogonal it first must be an nxn matrix and that $|AX|\ ∀X ∈ ℝ^n$ must be equal to $|X|$; along with $(A^t)(A) = I$ and the columns of $A$ form an orthonormal subset of $ℝ^n.$

keeping all of this in mind, what I first did was to just make sure and check to see if the magnitude of each column equaled $1$ and saw that they do indeed have a length of $1$, so that's good there. From here I took the dot product of each column and then saw that only $A_1$ dotted with $A_2$ works and have come to the conclusion that I need to change A_3, and this is where I'm stuck. Without doing what seems like endless trial and error, how does one solve this problem? I'm trying to keep all of these ideas in mind I'm just stuck in a rut it feels like. Thanks for the help!

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  • $\begingroup$ Find the orthogonal complement of the space spanned by the first two columns, or notice that their third coordinate is zero. $\endgroup$
    – amd
    Commented Apr 10, 2017 at 3:50

2 Answers 2

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Assume that last column is $$x=\left(\begin{array} ba \\ b \\ c \\ \end{array}\right).$$ Since third column is orthogonal to first two, we get $$3a/5-4b/5=0,$$ $$4a/5+3b/5=0$$ which implies that $a=0,b=0.$ Now magnitude of $x$ is $1$ gives that $c=1$. Hence we get $x$.

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  • $\begingroup$ thanks for the comment! My question is how/why does the third column being orthogonal to the first two result in you summing the elements in the first two columns multiplied by x? $\endgroup$ Commented Apr 10, 2017 at 20:54
  • $\begingroup$ Just the way you checked that "$A_1$ dotted with $A_2$ works". $\endgroup$ Commented Apr 11, 2017 at 13:56
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Think of the first two columns of the matrix as vectors that span the xy-plane. You already checked that they are orthonormal. Then clearly a vector that is orthogonal to those two vectors is one that is perpendicular to the xy-plane, i.e., points in the z direction, such as (0, 0, 1)$^\mathrm T.$ The other possible normal choice is (0, 0, –1)$^\mathrm T.$

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