- #1

squenshl

- 479

- 4

- 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.

$$c_1\mathbf{v}_1 = \mathbf{0}.$$

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.