I'm trying to understand the difference between Upper Triangular and Diagonal Matrices for square matrices. There is a theorem that says every matrix can be made into an upper triangular matrix. Now if my matrix as the same number of eigenvectors as its dimension then I can make a Diagonal matrix by changing the basis to the eigenspace. So in this case, why will I ever prefer to make a matrix Upper Triangular(and not diagonal) when I can make it diagonal?
Upper Triangular and Diagonal Matrices
abstract-algebradiagonalizationlinear algebramatrices
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I don't know whether you are dealing with $2\times 2$ matrices or general $n \times $n$ matrices. The result is true in either case.
It may not be clear to you what these spaces are. Define addition of matrices by adding corresponding entries. So for example $$\begin{bmatrix} 1 & 2\\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 5 & 3\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 5\\ 0 & 4 \end{bmatrix} $$ If $c$ is a constant (a scalar, a number) then you multiply a matrix by $c$ by multiplying each entry by $c$. So for example $$3\begin{bmatrix} 1 & 2\\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 6\\ 0 & 9 \end{bmatrix} $$
A vector space of matrices is a collection $V$ of matrices (of the same size) such that if $A$ and $B$ are matrices in the collection, then so is the sum $A+B$, and also if $c$ is any scalar, then $cA$ is in the collection.
So typically a vector space of matrices will have many matrices in it. The only vector space of matrices that consists of a single matrix is the space whose only element is the all $0$'s matrix.
In particular, the identity matrix by itself ($1$'s down the main diagonal, $0$'s elsewhere) is not a subspace of the collection of $2\times 2$ matrices, for if the identity matrix $I$ is in the subspace, then $cI$ has to be in the subspace for all numbers $c$. The collection of all matrices which are $0$ off diagonal, and have all diagonal terms equal is a subspace of the space of all matrices. Maybe that will take care of part of your objection.
Let $V$ be any vector space, and take a collection $U$ of some of the elements of $V$. Then $U$ is called a subspace of $V$ if $U$ by itself is a vector space, meaning that the sum of any two elements of $U$ is in $U$, and any constant times an element of $U$ is in $U$.
You quoted something to the effect that a certain $D$ is a subspace of the space of upper triangular matrices. That's not true. The collection of all matrices of the shape you described, with everything off diagonal equal to $0$, is a subspace. So $D$ is supposed to be not a single matrix, it is a largish collection of matrices.
Now let's look at your particular problem. Let $V$ be the collection of all upper triangular matrices. Is this a vector space? Take any two upper triangular matrices $A$ and $B$. Is $A+B$ upper triangular? Yes. If $c$ is a constant, and $A$ is upper triangular, is $cA$ upper triangular? Yes. So $V$ is a vector space.
Let $D$ be the collection of all diagonal matrices? Is this a vector space? Yes, the sum of two diagonal matrices is diagonal, a constant times a diagonal matrix is a diagonal matrix. $D$ is a subspace of the upper triangular matrices, because any diagonal matrix is in particular upper triangular, it is a special upper triangular matrix.
No, you tweak the eigenvectors of $T$ but still preserve $0<\langle e_1\rangle<\langle e_1,e_2\rangle<\dots<\mathbb{F}^n$.
For example: $M=\begin{bmatrix}-1\\&1\end{bmatrix}$ and you want to introduce a nonzero coefficient in the upper-right in $T$. You try something like $$ \begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}-1&0\\0&1\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix}^{-1} $$
Best Answer
Perhaps that you are missing the fact that every diagonal matrix is upper triangular too. So, if a matrix is diagonalizable, it is, by definition, similar to a diagonal matrix. Otherwise, it is not similar to a diagonal matrix, but it is still similar to an upper triangular one (at least over an algebraically closed field, such as $\mathbb C$).