# Showing a matrix A is diagonalisable

• squenshl
In summary, the conversation discusses proving the linear independence of a set of eigenvectors for a d x d matrix with distinct eigenvalues. It is shown that a single vector v1 is linearly independent, and the goal is to show that it is the only possible solution. Using the definition of eigenvalues and eigenvectors, the equation c1v1=0 is considered to prove this. It is concluded that v1 is linearly independent as required.
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.

squenshl said:
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?
squenshl said:
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.
squenshl said:
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.

## 1. What does it mean for a matrix to be diagonalisable?

When a matrix A is diagonalisable, it means that there exists a diagonal matrix D and an invertible matrix P such that A = PDP-1. In other words, the matrix A can be transformed into a diagonal matrix by a change of basis.

## 2. How do I determine if a matrix is diagonalisable?

A matrix A is diagonalisable if it has n linearly independent eigenvectors, where n is the size of the matrix. This means that the matrix must have n distinct eigenvalues. Additionally, if the algebraic multiplicity of each eigenvalue is equal to its geometric multiplicity, then the matrix is also diagonalisable.

## 3. What is the importance of diagonalisability in linear algebra?

Diagonalisability is important because it simplifies calculations involving the matrix A. Since the matrix can be transformed into a diagonal matrix, operations such as exponentiation and matrix multiplication become much easier. Additionally, diagonalisability allows us to find the inverse of the matrix A more easily.

## 4. Can a non-square matrix be diagonalisable?

No, a non-square matrix cannot be diagonalisable. The definition of diagonalisability requires the matrix to have the same number of rows and columns, which is not possible for a non-square matrix.

## 5. How do I show that a matrix is diagonalisable?

To show that a matrix A is diagonalisable, you can follow these steps:

• Find the eigenvalues of the matrix A.
• For each eigenvalue, find its corresponding eigenvector.
• If the matrix has n distinct eigenvalues and n linearly independent eigenvectors, then it is diagonalisable.
• If the algebraic multiplicity of each eigenvalue is equal to its geometric multiplicity, then the matrix is also diagonalisable.
• Finally, use the eigenvectors to construct the invertible matrix P and the diagonal matrix D such that A = PDP-1.

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