I'm currently studying Linear Algebra on my own, and I've run into some difficulties with eigenvectors:

In my book, the Matrix A is given as: \begin{equation*} A= \begin{bmatrix} 1 & 3 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & -2 \\ \end{bmatrix} \end{equation*} With the eigenvalues $4,-2,-2$ and bases for the eigenspaces ${(1,1,0),(1,-1,0),(0,0,1)}$ So far so good, but now it sais, that if we have a basis $B'$ made up of the three eigenvectors (same as the bases for the eigenspaces I assume) then $A'$ for $T$ (the linear transformation with standard matrix $A$)relative to the basis $B'$ is diagonal, however when I calculate $A$ relative to $B'$, so \begin{equation*} \begin{bmatrix} 1&1&0&1 & 3 & 0 \\ 1&-1&0&3 & 1 & 0 \\ 0&0&1&0 & 0 & -2 \\ \end{bmatrix} \end{equation*} And then simplify $B'$ to the Identity matrix I don't get \begin{equation*} A'= \begin{bmatrix} 4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \\ \end{bmatrix} \end{equation*} But rather \begin{equation*} A'= \begin{bmatrix} 2 & 2 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & -2 \\ \end{bmatrix} \end{equation*} I tried reversing the two, so $A B'$, I also tried multiplying the eigenvectors with their corresponding eigenvalue, but those didn't get me anywhere, let alone made sense.

Thank you for everyone who takes the time, if this is just a really stupid mistake, I'm sorry. Please consider that I'm in highschool so that my brain might not develope yet for this question.

  • $\begingroup$ Evidently you calculated $A'$ incorrectly. Nobody can help you find the error since you don't tell us how you did this calculation $\endgroup$ Aug 7 at 15:26

1 Answer 1


There's no need to make derogatory remarks about yourself. It's not going to do you any good to view yourself as having an undeveloped brain. You're learning and that's all that matters to anyone here who cares about helping you.

Now, you've introduced some notation in your question that needs to be further clarified. You need to try and explicitly write out what some of these objects are. In particular, $T$ is the linear map associated with the matrix $A$. If we multiply $A$ by the column vector $(x_1,x_2,x_3)$, then we obtain: $$T(x_1,x_2,x_3) = (x_1+3x_2,3x_1+x_2,-2x_3)$$

This is just a consequence of matrix multiplication and recognizing that if $x \in \mathbb{R}^3$, then $T(x) = Ax$. In fact, I placed your matrix into an eigenvalue calculator (I'm too lazy to calculate the eigenvalues manually) and that calculator tells me that your eigenvalues are actually $4$ and $-2$. That's quite likely what you meant to write. If not, you need to check your calculations again. Now, you have the eigenvectors: $$B' = \{v_1 = (1,1,0),v_2 = (1,-1,0), v_3 = (0,0,1)\}$$ What this means is that $T(v_1) = 4v_1$, $T(v_2) = -2v_2$ and $T(v_3)= -2v_3$. Now, you're correct that these vectors do form a basis for $\mathbb{R}^3$. In order to find the matrix relative to this basis, let's write the following: $$T(v_1) = 4v_1 +0v_2 + 0v_3$$ $$T(v_2) = 0v_1 -2v_2+0v_3$$ $$T(v_2) =0v_1 + 0v_2 -2v_3$$ By definition, the matrix of $T$ relative to $B'$ is going to be given by the following scheme: $$[T(v_1) \ T(v_2) \ T(v_3)]$$ where $T(v_1)$ just refers to the first column and so on. Moreover, if we are truly finding the matrix of $T$ relative to $B'$, then we need to represent $T(v_i)$ in terms of its expansion in the basis $B'$. In other words: $$T(v_1) = (4,0,0)$$ $$T(v_2) = (0,-2,0)$$ $$T(v_3) = (0,0,-2)$$ These are the coordinates of each $T(v_i)$ in terms of the basis $B'$. So, you actually do get a diagonal matrix in this case.

Let me explain the above a bit more, though I'm sure that your textbook will contain the exact definition. So, let $T: V \to W$ be a linear map with $V$ and $W$ being finite dimensional. Let $(v_1,\ldots,v_n)$ be a basis of $V$ and let $(w_1,\ldots,w_m)$ be a basis of $W$. Then, the matrix of $T$ relative to these bases is calculated as follows:

  1. Observe that each $T(v_j)$ is a vector in $W$. So, for each $i \in \{1,\ldots,n\}$, we can find scalars $(a_{ij})_{i=1}^{m}$ such that: $$T(v_j) = a_{1j} w_1 + a_{2j} w_2 + \ldots + a_{mj}w_m$$

In other words, in the basis $(w_1,\ldots,w_m)$, the coordinates of $T(v_j)$ are $(a_{1j},a_{2j},\ldots,a_{mj})$.

  1. Take the coordinates of each $T(v_j)$ with respect to the basis $(w_1,\ldots,w_m)$. Each of these becomes the column of a matrix. This matrix is, then, the matrix of $T$ relative to these two bases.

Once again, this definition should be in your book and it should've been thoroughly motivated, so you need to go back and have a look at it again.


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