MHB How Do I Navigate Change of Basis in Linear Transformations?

[ognik]

Hi, sadly my textbook assumes a knowledge I didn't have, of change of basis matrices & coordinate systems for linear transformations; so I have been trolling around the web to fill in the gaps as best I can.

I have an open post that has no replies - http://mathhelpboards.com/linear-abstract-algebra-14/follow-basis-question-16258.html - I would really appreciate if someone could at least check the 5 hesitant assertions I make in that post, even a yes/no would help please.

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I've moved on as best I can anyway, been working through an example from Khan Academy - but there is one aspect I just don't get.

The example transformation is a simple reflection through a plane V, the example sets out to show that changing the basis can make it easier to find the transformation matrix A. So we have $T(\vec{x})=A\vec{x}$, and an orthonormal basis in $R^3 , B=\left\{ v_1, v_2, v_3 \right\}$. If we allow $v_1, v_2 $ to span the plane V, then the transformation for these basis vectors would be $ T(\vec{v_1})= \vec{v_1}, T(\vec{v_2})= \vec{v_2}, T(\vec{v_3})= -\vec{v_3} $.

The example suggests that a change of basis matrix C comprised of the vectors of B is a 'natural' basis to use - 1st questions, is it always likely to be the best basis to use? We could use any orthonormal basis in $R^3$? (In this case it happens to simplify the transformation, what if it didn't?)

So I follow we want to find $T(\vec{x})=CDC^{-1} \vec{x}$ and from there, $A=CDC^T $

The example then says $ \vec{v_1}=1.\vec{v_1} + 0.\vec{v_2} +0.\vec{v_3}$ (ok), and then $ \left[ \vec{v_1} \right]_B = \begin{bmatrix}1\\0\\ 0 \end{bmatrix} etc.$

I'm just having a blank spot with this for some reason and maybe a different perspective on it would help me 'click'. I can see they have written $\vec{v}$ as a linear combination of the vectors in B, so the coordinates must be w.r.t. B.
But somehow we seem to have jumped from the vectors being in the standard basis (see $ T(\vec{v_1})= \vec{v_1}, T(\vec{v_2})= \vec{v_2}, T(\vec{v_3})= -\vec{v_3} $ above) - to those same vectors also being in B; for whatever reason, this just stops my mind at this point. (If I accept it, the rest of the example is easy to follow). Try to convince me please :-)

 

[Deveno]

As far as the vector space is concerned, ALL bases are equally good.

But we humans find some calculations more difficult than others: for example $37 + 0$ is easier to evaluate than $37 + 41$ (I can answer the first in a fraction (say half) of a second, the latter takes me 2-3 seconds, a sixfold decrease in efficiency).

Now matrices, being the beasts they are, can require a LOT of calculations to compute a matrix product (sums are not so bad). It would be nice if multiplying matrices were as easy as multiplying numbers (scalars).

One way this can happen, is if matrices are DIAGONAL. Now if it were to happen that the matrix of a linear transformation was diagonal, calculation with that matrix (in that basis) would be a LOT easier. It turns out that this isn't always possible, and to determine when it IS possible, we need to study things called eigenvectors.

So, although the Khan Academy video isn't really explicitly saying as much, what you have in the reflection example is a case where the basis vectors are chosen to be eigenvectors.

Indeed, vectors in the plane of reflection are all eigenvectors with a eigenvalue of $1$, this is to say they are left unchanged by the reflection. The $v_3$ is then a basis for what is called the orthogonal complement of the plane, which can be shown to be "independent" of the plane in the sense that:

$\Bbb R^3 = P \oplus \langle v_3\rangle$.

I do not know if your text has covered direct sums or complementary subspaces, yet.

In any case, if $B = \{v_1,v_2,v_3\}$, then:

$[T]_B = \begin{bmatrix}1&0&0\\0&1&0\\0&0&-1\end{bmatrix}$

which is indeed a diagonal matrix, and easy for us simple humans to do calculations with.

Is this the "best" way to represent $T$? That's not really the right question-it's a CANONICAL way to express $T$, in what is known as JORDAN NORMAL FORM. $B$ forms an eigenbasis for $\Bbb R^3$ with respect to $T$, which turns out to be important for many APPLICATIONS of linear algebra (where calculation difficulty matters more).

One small point: if $Tv = \lambda v$ for some scalar $\lambda$, this is true no matter WHICH basis we express $T$ and $v$ in:

Standard basis- using the matrix $A$ for $T$:

$Tv = Av = \lambda v$.

Another basis - using the matrix $P^{-1}AP$ for $T$:

$[Tv]_B = P^{1}AP[v]_B = P^{-1}A(v) = P^{-1}(\lambda v) = \lambda P^{-1}v = \lambda[v]_B = [\lambda v]_B$

so the apparent "switching back and forth" between bases in the videos doesn't affect the equations:

$T(v_1) = v_1,\ T(v_2) = v_2$, and $T(v_3) = -v_3$.

 

[ognik]

Yes indeed, the next section from Khan was eigenvectors, so you correctly anticipated that. I've worked through eigenvectors now, but it doesn't help with the blind spot I mentioned - I can find the diagonal matrix using eigenvalues, but I just can't see/accept how it was done without eifenvalues, ie: $ \vec{v_1}=1.\vec{v_1} + 0.\vec{v_2} +0.\vec{v_3}$ (ok), and then $ \left[ \vec{v_1} \right]_B = \begin{bmatrix}1\\0\\ 0 \end{bmatrix} etc.$

It seems to me this somehow jumps those vectors from one basis to another, using itself...

 

[Deveno]

In the basis $B = \{v_1,v_2,v_3\}$ we have $[v_1]_B = (1,0,0)$ (these are the $B$-coordinates of $v_1$, which in another post I said I like to denote as $[1,0,0]_B$ so as to avoid confusion with $e_1$).

The argument that $[T]_B([v_1]_B) = [Tv_1]_B = [v_1]_B$ is based on GEOMETRY, reflecting 3-space through a plane leaves the plane itself unchanged. In the basis $B$, therefore, the first column of the matrix for $T$ is $(1,0,0)^t$.

Just to be clear- a triple of numbers (which we associate naturally with a POINT in $\Bbb R^3$) is just a triple of numbers:

$(3,2,1)$ might mean: 3 feet east, 2 feet north, and 1 foot up, or it might mean: 3 miles east, 2 yards north and 1 inch up.

For example, you might ask me how far I traveled today, and I might answer "4". You should naturally wonder-"four..what?". Four miles, kilometers, city blocks, yards? The standard basis is, effectively, using whatever "unit" $1$ is standing for in the real numbers, and applying the SAME SCALE orthogonally to each dimension. Other scales, and "skewings" are possible, and perhaps more convenient. We get to choose a basis as it suits us.

The points in the plane on the line $y = ax$ have a simpler form if we rotate by the angle $\arctan(a)$. This doesn't change the "line-ness" of the points on the line, but it may lead to less messy calculations. With a lot of linear transformations, finding a "simpler" form of the matrix (by choosing the right basis) is desirable.

 

[johng1]

Hi,
Maybe your "blind" spot is geometrical in nature. First think in 2 dimensions (it's easier to draw pictures here). Consider the problem of reflection of points about the x-axis; i.e. a point $(x,y)\rightarrow (x,-y)$. The matrix of this transformation (with respect to the standard basis) is clearly
$$\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$
Now in three dimensions, the problem of reflecting points about the x-y plane is entirely similar. A point $(x,y,z)\rightarrow(x,y,-z)$. The matrix of this transformation, again with respect to the standard basis, is
$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&-1\end{bmatrix}$$

 

[ognik]

Thanks guys, always appreciate the patience.

So $ \left[ v_1 \right]_B = (1,0,0) $; is $ v_1 \:also\: = (1,0,0)? $ ie. w.r.t. the standard basis.

 

[Deveno]

Thanks guys, always appreciate the patience.

So $ \left[ v_1 \right]_B = (1,0,0) $; is $ v_1 \:also\: = (1,0,0)? $ ie. w.r.t. the standard basis.

Only if $v_1 = e_1$. What $v_1$ is in the standard basis will depend, in general, on WHICH plane we are reflecting through, if we were, for example, reflecting in the $xy$-plane, then we would have $v_1 = e_1$.

The point of the approach taken by Khan Academy, is that GIVEN the plane, we can presumably FIND $v_1,v_2$ that span it. Unless the plane contains the $x$-axis as a subspace, $e_1$ (the vector (1,0,0) in the standard basis) will not be one of those spanning vectors.

(N.B. I am using the "usual" identification of $e_1$ with the $x$-axis unit vector (also denoted $\mathbf{i}$), $e_2$ with the $y$-axis unit vector (sim. $\mathbf{j}$), and $e_3$ with the $z$-axis unit vector (aka $\mathbf{k}$). This identification is COMPLETELY ARBITRARY, and should not be taken as gospel).

 

[ognik]

In the basis $B = \{v_1,v_2,v_3\}$ we have $[v_1]_B = (1,0,0)$

Sorry, how do we know that?

 

[Deveno]

Because:

$v_1 = 1v_2 + 0v_2 + 0v_3$ and since $B$ is a basis, this is the UNIQUE (the only one!) way to write $v_1$ as a linear combination of the $v_j$.

If we had ANY OTHER WAY, say:

$v_1 = av_1 + bv_2 + cv_3$, then:

$0 = v_1 - v_1 = (av_1 + bv_2 + cv_3) - (1v_1 + 0v_2 + 0v_3) = (a-1)v_1 + (b-0)v_2 + (c-0)v_3$

$= (a-1)v_1 + bv_2 + cv_3$.

By the linear independence of $v_1,v_2,v_3$ (since they are a basis), this means:

$a - 1 = b = c = 0$, which means our "two ways" are really both the "same way".

 

[ognik]

Thanks all, I believe I have sorted out the muddle in my head - just a temporary inability to see the obvious :-)