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suppose i have a linear trasformation $$T(X)=\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}X,T:M_{2}(\mathbb{R})\rightarrow M_{2}(\mathbb{R})$$ and i want to find the eigenvalues, i started by calculating the transformation of the basis matrices. $$T\left(\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right)=\left(\begin{array}{cc} 1 & 1\\ 1 & 1 \end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right)=\left(\begin{array}{cc} 1 & 0\\ 1 & 0 \end{array}\right)$$ and did it for the rest of the basis vectors. now, when i build the transformation matrix does it matter if i organize them by row or by column? i mean is there a difference between $$\left[T\right]_{E}=\left(\begin{array}{cccc} 1 & 0 & 1 & 0\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 0 & 1 & 0 & 1 \end{array}\right)%% %% \left[T\right]_{E}=\left(\begin{array}{cccc} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1 \end{array}\right)$$

english is not my main language so i had some hard time figuring out what to google in order to find a solution.

edit: i have the matrices $$\left(\begin{array}{cc} 1 & 0\\ 1 & 0 \end{array}\right) \left(\begin{array}{cc} 0 & 1\\ 0 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & 0\\ 1 & 0 \end{array}\right) \left(\begin{array}{cc} 0 & 1\\ 0 & 1 \end{array}\right) $$ and the problem is how to order their entries: as rows or columns?

thank you!

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  • $\begingroup$ Well, are you flattening elements of $M_2(\mathbb R)$ into column vectors or row vectors? The answer to that determines how you should organize the matrix. $\endgroup$
    – amd
    May 4, 2020 at 19:41

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If I understand, you’re considering whether to order your basis for $M_2$ as $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, $$ or $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 &0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} .$$ If so, then this won’t matter. It just amounts to a change of basis, which won’t alter eigenvalues.

EDIT: If we have a linear transformation $T : V \to V$, where $V$ is a $d$-dimensional vector space, we can encode it as a square matrix as follows. Let $\{v_1, \ldots, v_d \}$ be a basis for $V$. Define a $d \times d$ matrix $A = [a_{i, j}]_{i, j = 1}^d$ by looking at $T v_j$ and writing $T v_j = a_{1, j} v_1 + \cdots + a_{d, j} v_d = \sum_{i = 1}^d a_{i, j} v_j$. The expression is unique because $\{ v_1, \ldots, v_d \}$ is a basis for $V$. This means that the $j$th column of the matrix $A$ depicts how $T$ acts on $v_j$. It is, however, vital that we keep the order of the basis consistent when calculating eigenvalues, i.e. we can't swap them around.

So if we were to take our basis to be \begin{align*} v_1 & = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} , & v_2 & = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \\ v_3 & = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, & v_4 & = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} , \end{align*} then we'd compute our matrix as follows. \begin{align*} T v_1 & = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \\ & = v_1 + v_3 , \\ T v_2 & = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} \\ & = v_2 + v_4, \\ T v_3 & = v_1 + v_3 , \\ T v_4 & = v_2 + v_4 . \end{align*} The matrix for $T$ under this basis would then be $$A = \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{bmatrix} .$$

Now for fun, let's reorder the basis, and consider the basis \begin{align*} w_1 & = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, & w_2 & = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} , \\ w_3 & = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, & w_4 & = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \end{align*} This is how I interpreted your question about "rows" and "columns", I'm sorry if I misunderstood you. If we compute the operator on this basis, we'll get \begin{align*} T w_1 & = w_1 + w_2, & T w_2 & = w_1 + w_2, \\ T w_3 & = w_3 + w_4, & T w_4 & = w_3 + w_4 , \end{align*} yielding a matrix $$\begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix} .$$

As far as computing eigenvalues, these matrices are both just as good as each other. They're what we call equivalent, meaning that there exists an invertible $d \times d$ matrix $S$ such that $A = S B S^{-1}$. In general, if we take some linear transformation $T : V \to V$, and we take any two bases of $V$, say $\{ v_1, \ldots, v_d \}, \{ w_1, \ldots, w_d\}$, and we write the matrix $A$ for $T$ based on $\{ v_1, \ldots, v_d\}$, and the matrix $B$ for $T$ based on $\{ w_1, \ldots, w_d \}$, then $A$ and $B$ will be equivalent. Any two equivalent matrices will have the same characteristic polynomials, and thus the same eigenvalues.

I also wanna comment that this is specifically the way you build these square matrices when $T$ maps from a vector space $V$ to itself. In general, we'll have linear transformations $T : V \to W$, where we obviously can't use one basis to compute $T$. Instead we'll need a basis for $V$ and a basis for $W$. For that, you might look at Section 1.4 of these notes.

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  • $\begingroup$ i edited the question, is it still the same? $\endgroup$ May 4, 2020 at 14:28
  • $\begingroup$ @EladElmakias I edited my answer, does it address your question now? $\endgroup$
    – AJY
    May 4, 2020 at 22:13
  • $\begingroup$ first of all thank you but no thats not what i meant, when you placed the Tv1 ... Tv4 in the matrix you took each row of the answer and placed it in the column of [T]A and what i did at first is taking each column of each answer and put it in each column of [T]A. in my edited answer you can see 2 options of the final answer, one of them is identical to yours and the other one is not. what im having troubles to understand is why does it matter which method I do? $\endgroup$ May 5, 2020 at 15:21
  • $\begingroup$ @EladElmakias It cooperates with matrix multiplication. Suppose I had linear maps $S: V \to W, T: W \to X$, and I had bases $\{v_1, \ldots, v_m\}, \{w_1, \ldots, w_m\}, \{x_1, \ldots, x_k\}$ for the respective spaces. If I made a matrix $A$ for $S$ with those bases for $V, W$, and a matrix $B$ for $T$ with the bases for $W, X$, then the matrix of $T \circ S$ would be $BA$. $\endgroup$
    – AJY
    May 5, 2020 at 17:59
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    $\begingroup$ The only problem i have is when i try to take a matrix, turn it into a vector and insert it to another matrix, the problem is how i create this vector? By taking each row of the matrix or each column $\endgroup$ May 5, 2020 at 18:08

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