Prove but \(T\) is diagonalizable if and only if the matrix of \(T\) is similar to a diagonal matrix.

Try 2.

Given similarity:

So we have that:

\begin{equation} D = S^{-1} A S \end{equation}

where, \(D\) is diagonal. We apply \(S\) to both sides to yield:

\begin{equation} SD = AS \end{equation}

Now, note that \(S\) is invertible. This means that its column s are linearly independent (as it is an operator, which means it is injective, and hence has a zero null space; that indicates that the dimension of its range is that of the whole space: indicating its columns vectors are spanning; there is \(dim\ V\) such columns, so it is a basis and hence linearly independent).

Two Variables

Let’s begin with the equations:

\begin{equation} \begin{cases} 2x+y = 3 \\ x - y = 0 \end{cases} \end{equation}

We will first change this into a matrix equation:

\begin{equation} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix} \end{equation}

We need to find, then, the inverse of:

\begin{equation} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \end{equation}

Suppose \(\mathbb{F} = \mathbb{R}\), and \(V \neq \{0\}\). Replace the positivity condition with the condition that \(\langle v,v \rangle > 0\) for some \(v \in V\). Show that this change in definition does not change the set of functions from \(V \times V\) to \(\mathbb{R}\) that are inner products on \(V\).

We hope to show that \(\langle v,v \rangle >0\) for some \(v \in V\) implies that \(\langle v,v \rangle \geq 0\) for all \(v \in V\) in real vector spaces.

Proof: identity of a group is unique

Assume for contradiction that there exists two identities \(e_1\) and \(e_2\) which are identities of the group \(A\). Take also an \(a \in A\).

Given both \(e_1\) and \(e_2\) are identities, we have that:

\begin{equation} a * e_1 = a \end{equation}

as well as

\begin{equation} a * e_2 = a \end{equation}

Therefore, we have by the transitive property that:

\begin{equation} a * e_1 = a*e_2 \end{equation}

We declare known battery voltage \(E(t)\).

Here are the \(y\) values.

\begin{equation} \begin{cases} \dv{x_1}{t} = y_{4}\\ \dv{x_2}{t} = y_{3}\\ \dv{x_3}{t} = y_{1}\\ \dv{x_4}{t} = y_{2}\\ \end{cases} \end{equation}

And here are some of the \(x\) values.

\begin{equation} \begin{cases} \dv{x_4}{t}=-\frac{2}{RC}x_2-\frac{1}{RC}x_{3}-\frac{2E(t)}{R} \\ \dv{y_1}{t}=-\frac{1}{LC}x_2-\frac{E(t)}{C} \\ \dv{y_4}{t} = -\frac{R}{L}y_2-\frac{2E(t)}{L} \end{cases} \end{equation}

Right off the bat, we can see that we can make one substitution. That, given:

\begin{equation} \begin{cases} \dv{x_4}{t}=-\frac{2}{RC}x_2-\frac{1}{RC}x_{3}-\frac{2E(t)}{R} \\ \dv{x_4}{t} = y_{2} \end{cases} \end{equation}