Why is it that in a transformation, if the eigenvectors are chosen as the basis vectors, the transformation matrix $A$ would turn out to be that diagonal eigenvalues matrix $\Lambda$ and the transformation becomes $\Lambda \vec{x}=\vec{c}$ ?

What is the reason or cause that the transformation matrix could easily just be the eigenvalues matrix $\Lambda$?

I tried to prove it to show myself but somehow, it doesn't look very right. Let $\vec{b_i}$ be the eigenvectors that form the basis for the vector $\vec{v}$ and the $c_n$ are combination to the eigenvectors...

$\vec{v} = c_1 \vec{b_1} + c_2 \vec{b_2} + ... + c_n \vec{b_n} $

$ T(\vec{v})= c_1 T(\vec{b_1}) + c_2 T(\vec{b_2}) + ... + c_n T(\vec{b_n}) $

$ T(\vec{v})= c_1 A\vec{b_1} + c_2 A\vec{b_2} + ... + c_n A\vec{b_n} $

$ T(\vec{v})= c_1 \lambda_1 \vec{b_1} + c_2 \lambda_2\vec{b_2} + ... + c_n \lambda_n\vec{b_n} $

$ T(\vec{v})= \begin{bmatrix} & & \\ \vec{b_1} & \vec{b_2} &...& \vec{b_n}\\ & & \end{bmatrix} \begin{bmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_n \end{bmatrix} \begin{bmatrix} c_1 & & & \\ & c_2 & & \\ & & \ddots & \\ & & & c_n \end{bmatrix} $

The last line is wrong but I'm trying to force out that $\Lambda$ which I couldn't.


Let me back up a bit. Remember what the point is of the "coordinate matrix of a linear transformation with respect to a basis."

If $T\colon\mathbf{V}\to\mathbf{V}$ is a linear transformation, and $\beta=[\mathbf{v}_1,\ldots,\mathbf{v}_n]$ is a basis for $T$, then the values of $T$ at every vector in $\mathbf{V}$ is completely determined by its values on $\mathbf{v}_1,\ldots,\mathbf{v}_n$. Why? Because given any $\mathbf{x}\in\mathbf{V}$, we know there exist (unique) scalars $c_1,\ldots,c_n$ such that $\mathbf{x}=c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n$, so $$T(\mathbf{x}) = T(c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n) = c_1T(\mathbf{v}_1)+\cdots+c_nT(\mathbf{v}_n).$$ So if you know what $T(\mathbf{v}_1),\ldots,T(\mathbf{v}_n)$ are, you know that $T(\mathbf{x})$ is for all $\mathbf{x}\in\mathbf{V}$.

Now, since $\beta$ is a basis, each $T(\mathbf{v}_i)$ can be expressed as a linear combination of the $\mathbf{v}_i$ in a unique way; that is, there are scalars $a_{11}, a_{12},\ldots,a_{nn}$ such that $$\begin{align*} T(\mathbf{v}_1) &= a_{11}\mathbf{v}_1 + a_{21}\mathbf{v}_2 + \cdots +a_{n1}\mathbf{v}_n\\ T(\mathbf{v}_2) &= a_{12}\mathbf{v}_1 + a_{22}\mathbf{v}_2 + \cdots + a_{n2}\mathbf{v}_n\\ &\vdots\\ T(\mathbf{v}_n) &= a_{1n}\mathbf{v}_1 + a_{2n}\mathbf{v}_2 + \cdots + a_{nn}\mathbf{v}_n. \end{align*}$$ Then $$\begin{align*} G(\mathbf{x}) &= c_1T(\mathbf{v}_1 + \cdots + c_nT(\mathbf{v}_n)\\ &= c_1\left(a_{11}\mathbf{v}_1 + a_{21}\mathbf{v}_2 + \cdots +a_{n1}\mathbf{v}_n\right)\\ &\qquad+c_2\left(a_{12}\mathbf{v}_1 + a_{22}\mathbf{v}_2 + \cdots + a_{n2}\mathbf{v}_n\right)\\ &\qquad+\cdots+c_n\left(a_{1n}\mathbf{v}_1 + a_{2n}\mathbf{v}_2 + \cdots + a_{nn}\mathbf{v}_n\right)\\ &= \Bigl(c_1a_{11} + c_2a_{12} + \cdots + c_na_{1n}\Bigr)\mathbf{v}_1\\ &\qquad +\Bigl(c_1a_{21} + c_2a_{22} + \cdots + c_na_{2n}\Bigr)\mathbf{v}_2\\ &\qquad+\cdots+ \Bigl(c_1a_{n1}+c_2a_{n2}+\cdots+c_na_{nn}\Bigr)\mathbf{v}_n\\ &= \left(\begin{array}{cccc} \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_n\end{array}\right) \left(\begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{nn}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{array}\right) \left(\begin{array}{c}c_1\\c_2\\\vdots\\c_n\end{array}\right). \end{align*}$$

The column vector $$\left(\begin{array}{c}c_1\\c_2\\\vdots\\c_n\end{array}\right)$$ represents the vector $\mathbf{x}$; it is the "coordinate vector of $\mathbf{x}$ with respect to $\beta$". The matrix $$\left(\begin{array}{ccc} a_{11} & \cdots & a_{1n}\\ \vdots & \ddots & \vdots\\ a_{n1} & \cdots & a_{nn}\end{array}\right)$$ is the matrix that "codifies" the images of $\mathbf{v}_i$ under $T$ in terms of $\beta$: the $i$th column is the coordinate vector of $T(\mathbf{v}_i)$ relative to $\beta$. This is the "coordinate matrix of $T$ relative to $\beta$."

What happens if $\mathbf{v}_1$ is an eigenvector corresponding to $\lambda_1$? Then $$T(\mathbf{v}_1) = \lambda_1\mathbf{v}_1 = \lambda_1\mathbf{v}_1 + 0\mathbf{v}_2 + 0\mathbf{v}_3 + \cdots + 0\mathbf{v}_n.$$ That is, the first column of the coordinate matrix is just $$\begin{array}{c} \lambda_1 \\ 0 \\ 0\\ \vdots \\ 0\\ 0\end{array}$$ What if $\mathbf{v}_2$ is an eigenvector corresponding to $\lambda_2$? Then $T(\mathbf{v}_2)=\lambda_2\mathbf{v}_2$. To write it as a linear combination of $\beta$, we just put $\lambda_2$ for the coefficient of $\mathbf{v}_2$, and $0$s for every other coefficient: $$T(\mathbf{v}_2) = \lambda_2\mathbf{v}_2 = 0\mathbf{v}_1 + \lambda_2\mathbf{v}_2 + 0\mathbf{v}_3 + \cdots + 0\mathbf{v}_n.$$ So the second column of the coordinate matrix is just $$\begin{array}{c}0\\\lambda_2\\0\\\vdots\\0\\0\end{array}$$ And so on: if $\mathbf{v}_i$ is an eigenvector corresponding to $\lambda_i$, then $$T(\mathbf{v}_i) = \lambda_i\mathbf{v}_i = 0\mathbf{v}_1+0\mathbf{v}_2+\cdots+0\mathbf{v}_{i-1} + \lambda_i\mathbf{v}_i + 0\mathbf{v}_{i+1} + \cdots+0\mathbf{v}_n,$$ so the $i$th column of the coordinate matrix will have $0$s everywhere except in the $i$th row, where it will have $\lambda_i$.

And so, if the entire basis is made up of eigenvectors, then the coordinate matrix will be such that the $i$th column has zeros everywhere except (perhaps) in the $i$th row, where it has the eigenvalue of $\mathbf{v}_i$; that is, we get a diagonal matrix $$\Lambda = \left(\begin{array}{cccc} \lambda_1 & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n \end{array}\right).$$ And that is the coordinate matrix of $T$ with respect to $\beta$.

  • $\begingroup$ Thanks for such a clear explanation. So can I say that for $T([\vec{x}]_v)$, that's the input $\vec{x}$ with respect to the eigenvector basis $V$, if the output was: $$ T([\vec{x}]_v) = \begin{bmatrix} \vec{v_1} & \vec{v_1} & ... & \vec{v_1} \end{bmatrix} \left[\begin{array}{cccc} \lambda_1 & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n \end{array}\right] \left[\begin{array}{c}c_1\\c_2\\\vdots\\c_n\end{array}\right] $$ This means that the output coordinates are with respect to the standard basis, is this right? $\endgroup$ – xenon Aug 20 '11 at 7:32
  • $\begingroup$ And if I wanted the output coordinates with respect to the eigenvectors basis, then it would be just: $$T([\vec{x}]_v) = \left[\begin{array}{cccc} \lambda_1 & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n \end{array}\right] \left[\begin{array}{c}c_1\\c_2\\\vdots\\c_n\end{array}\right]$$ Is this right too? $\endgroup$ – xenon Aug 20 '11 at 7:32
  • $\begingroup$ @xEnOn: Careful: $T$ is an "abstract linear transformation". It acts on vectors in the vector space. $[\vec{x}]_v$ represents the coordinate vector of $\vec{x}$ relative to $v$. $T$ would "misunderstand" $[\vec{x}]_v$ if you try feeding it. What you need is the coordinate matrix of $T$. That is, you need $[T]_v^v$ applied to $[\vec{x}]_v$, and that will give you $[T(\vec{x})]_v$. I.e., $[T]_v^v([\vec{x}]_v) = [T(x)]_v$. $\endgroup$ – Arturo Magidin Aug 20 '11 at 16:35
  • $\begingroup$ Thank you so much for your help! :) $\endgroup$ – xenon Aug 21 '11 at 2:01

The matrix of a linear transformation with respect to a basis has as columns the coordinates of the images of the basis elements under the transformation. If you have a basis of eigenvectors then the corresponding matrix will be diagonal because $A v_i = \lambda_i v_i$, that is, there is only one non-zero coordinate (more precisely, at most one, because $\lambda_i$ may be $0$).

  • 1
    $\begingroup$ A way of seeing this same answer more intuitively is to think about a diagonal matrix as a linear transformation that corresponds to a scaling of the basis vectors. The eigenvectors are exactly the ones that are (purely) scaled by a given linear transformation, so if we use these as a basis then the matrix is diagonal. $\endgroup$ – Vhailor Aug 19 '11 at 15:46
  • $\begingroup$ Thanks. But I'm still a little confuse. I have updated my question with a some lines that I did to try to prove to myself but somehow, I still feel I'm missing out something. In the end, I couldn't eventually form into the expected $T(\vec{v})=\Lambda \vec{v}$. $\endgroup$ – xenon Aug 19 '11 at 16:22
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    $\begingroup$ @xEnOn: Remember that linear transformations act on vectors, but matrices act on the coordinates. In your example the coordinates of $\vec{v}$ are the (column) vector $(c_1,c_2,\ldots,c_n)^T$ ($T$ stands for transpose. OTOH the coordinates of the transformed vector $T(\vec{v})$ are $(\lambda_1c_1,\lambda_2c_2,\ldots,\lambda_nc_n)^T$. These are, indeed, related by $\Lambda$ as $$\left(\begin{array}{c} \lambda_1c_1 \\ \lambda_2c_2 \\ \vdots \\ \lambda_nc_n\end{array}\right)=\Lambda \left(\begin{array}{c} c_1 \\ c_2 \\ \vdots \\ c_n\end{array}\right).$$ $\endgroup$ – Jyrki Lahtonen Aug 19 '11 at 20:07

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