Showing a matrix A is diagonalisable

[squenshl]

Homework Statement

Here we show that if a $d\times d$ matrix $A$ has distinct eigenvalues $\lambda_1,\ldots,\lambda_d$ with eigenvectors $\mathbf{v}_1,\ldots,\mathbf{v}_d$, then it is diagonalisable, i.e., $\mathbb{R}^d$ has a basis of eigenvectors.

My question is suppose that $\{\mathbf{v}_1,\ldots,\mathbf{v}_n\}$, $n < d$ is linearly independent, and show that $\{\mathbf{v}_1,\ldots,\mathbf{v}_{n+1}\}$ is linearly independent.

Relevant Equations

None

Show that $\{\mathbf{v}_1\}$ is linearly independent. Simple enough let's consider
$$c_1\mathbf{v}_1 = \mathbf{0}.$$
Our goal is to show that $c_1 = 0$. By the definition of eigenvalues and eigenvectors we have $A\mathbf{v}_1= \lambda_1\mathbf{v}_1$. Let's multiply the above equation and $A$ to get
$$\mathbf{0} = A\times \mathbf{0} = A(c_1\mathbf{v}_1) = c_1A\mathbf{v}_1 = c_1\lambda_1\mathbf{v}_1.$$
An eigenvector is by definition a nonzero vector, and hence $\mathbf{v}_1\neq 0$. Thus, we must have $c_1\lambda_1 = 0$. Since $\lambda_1$ is distinct, we must have $c_1 = 0$. Hence, $\{\mathbf{v}_1\}$ is linearly independent as required.

 

[Mark44]

Homework Statement: Here we show that if a $d\times d$ matrix $A$ has distinct eigenvalues $\lambda_1,\ldots,\lambda_d$ with eigenvectors $\mathbf{v}_1,\ldots,\mathbf{v}_d$, then it is diagonalisable, i.e., $\mathbb{R}^d$ has a basis of eigenvectors.

My question is suppose that $\{\mathbf{v}_1,\ldots,\mathbf{v}_n\}$, $n < d$ is linearly independent, and show that $\{\mathbf{v}_1,\ldots,\mathbf{v}_{n+1}\}$ is linearly independent.
Homework Equations: None

Show that $\{\mathbf{v}_1\}$ is linearly independent.

This doesn't make sense to me. Generally when you're talking about linear dependence/independence, you're considering a set of two or more vectors, not just a single vector. It's almost trivial to prove that a single nonzero vector is a linearly independent set, just using the definition of lin. independence, but what's the point?

Simple enough let's consider
$$c_1\mathbf{v}_1 = \mathbf{0}.$$
Our goal is to show that $c_1 = 0$.

No it is not. The goal is to show that $c_1 = 0$ is the only possible solution. For example, if $v_1 = <1, 2>$ and $v_2 = <2, 4>$, then the equation $c_1v_1 + c_2v_2 = 0$ is a true statement if $c_1 = c_2 = 0$, but the two vectors are linearly dependent.

By the definition of eigenvalues and eigenvectors we have $A\mathbf{v}_1= \lambda_1\mathbf{v}_1$. Let's multiply the above equation and $A$ to get
$$\mathbf{0} = A\times \mathbf{0} = A(c_1\mathbf{v}_1) = c_1A\mathbf{v}_1 = c_1\lambda_1\mathbf{v}_1.$$
An eigenvector is by definition a nonzero vector, and hence $\mathbf{v}_1\neq 0$. Thus, we must have $c_1\lambda_1 = 0$. Since $\lambda_1$ is distinct, we must have $c_1 = 0$. Hence, $\{\mathbf{v}_1\}$ is linearly independent as required.