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Let $I$ be a 3-by-3 identity matrix and $\hat n$ be a unit vector orthonormal to some surface. What then does $I-\hat n\hat n$ mean geometrically? Also what does it mean to multiply this matrix/operator by a vector $v$? Some sort of projection?

Also, I don't understand what $\hat n\hat n$ means. Dimensionally it should be a 3-by-3 matrix because $I$ is such, right? 

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    Is $\hat{n}$ a $3\times 1$ or $1\times 3$?2012-07-01
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    @J.D. I am guessing it is a column vector, since vectors tend to be such, but it isn't specified? Does ot make a difference?2012-07-01
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    The only way I can think of for it to make some sense is to look at it as the scalar product times identity matrix. But then its zero, so it still doesn't make much sense. Context?2012-07-01
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    If $n$ is a column vector (as usual in most linear algebra textbooks), then $\hat n \hat n$ would probably mean $\hat n\ \hat n^{T},$ which is the projection in the direction of $\hat n.$2012-07-01
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    @J.D. Thanks for the insight!2012-07-01
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    @theodoreA please check the image in my updated answer.2012-07-01

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If $n$ is a column vector (as usual in most linear algebra textbooks), then $\hat n\hat n$ would probably mean $\hat n \hat n^{T},$ which is the projection in the direction of $\hat n.$

And $(I - \hat n \hat n) v$ is the vector that resembles the projection vector from the tip of $v$ into the plane. You can see it if you expand $(I - \hat n \hat n) v$. You get $v - v_n$ where $v_n$ is the projection of $v$ onto the direction of $n.$

You can see in the following image: $v - v_n = e$

enter image description here