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I'm teaching this topic to Year 12 students at present and have come across the following which I don't fully understand. I wonder if anyone can enlighten me as to the significance or otherwise of the following.

If three linearly dependent vectors (a, b & c) are used to construct a 3 x 3 matrix (M), so that they form the rows of the matrix.

If: αabc=0 (i.e. the liner dependence property)

Then the M will transform all points to a plane such that the vector <α,β,γ> is normal to the plane. I understand why such a matrix must transform to a plane since M's zero determinant results in a destruction of any volume in the transformation, however, I can't see why the normal should be related to the linear dependence property in this way?

Thanks for any help, particularly if couched in terms I can explain to Year 12 pupils.

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    You’re going to need another condition. The claim isn’t true if the rank of $M$ is less than two.2017-02-08
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    @amd Good point; the claim is only true if we interpret the phrase "all points to a plane" in a sufficiently generous manner.2017-02-08

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Let $\mathbf x$ be any point, i.e. a column vector, and let your notation $\langle\alpha,\beta,\gamma\rangle$ represent a row vector. Now consider the product $\langle\alpha,\beta,\gamma\rangle M \mathbf x$. Since matrix multiplication is associative, we have $$\left(\langle\alpha,\beta,\gamma\rangle M\right) \mathbf x = \langle\alpha,\beta,\gamma\rangle\left(M \mathbf x\right).$$ How do you interpret the left and right sides of this equation?

Edit: In case this wasn't clear: Start on the left-hand side. Expanding out the definition of matrix multiplication, we can rewrite $\langle\alpha,\beta,\gamma\rangle M=\alpha\mathbf a+\beta\mathbf b+\gamma\mathbf c$. But by the linear dependence assumption, that equals zero, i.e. the zero vector! So the above equation reduces to: $$0= \langle\alpha,\beta,\gamma\rangle\left(M \mathbf x\right).$$ Now on the right-hand side, $M\mathbf x$ is the point to which $M$ transforms $\mathbf x$. This equation states that its dot product with $\langle\alpha,\beta,\gamma\rangle$ is zero, which is another way to say that $M\mathbf x$ is orthogonal to $\langle\alpha,\beta,\gamma\rangle$. And the set of all vectors orthogonal to $\langle\alpha,\beta,\gamma\rangle$ is precisely the plane normal to $\langle\alpha,\beta,\gamma\rangle$.

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    Thanks for this, I take it this shows the vector equation of a plane being created with the normal as specified. Is there any physical significance to this fact (with the limitations suggested in the comments)?2017-02-08
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    @JonMannering This is a special case of $\ker(M^T)=(\operatorname{im}M)^\perp$.2017-02-09