# Proving that a property of linear independence in linear algebra

Given that $$x^{(0)}=\beta_{2}v^{(2)}+\beta_{3}v^{(3)}+\dots+\beta_{n}v^{(n)}$$ where $v^{(1)},v^{(2)},\dots,v^{(n)}$ are linearly independent eigenvectors and $\lambda_{1},\lambda_{2},\dots,\lambda_{n}$ are the corresponding eigenvalues.

(a) Show that for any vector $x=\sum_{i=1}^n \beta_{i}v^{(i)}$, the vector $x^{(0)}=(A-\lambda_{1}I)x$ satisfies the expression above.

(b) Find $\lambda_{3}$ using $x^{(0)}=(A-\lambda_{2}I)(A-\lambda_{1}I)x.$

What I tried

By substituting the vector $x^{(0)}=(A-\lambda_{1}I)x$ to the LHS of the expression above I realise there is the missing $\beta_{1}v^{(1)}$ term in the RHS of the expression and to account for the missing term, I believe it must have something to do with the term $(A-\lambda_{1}I)x$ which is the characterstic polynomial of the matrix $A$. Could anyone explain this to me, please?

• The RHS is not a polynomial expression. It is a square matrix multiplied with a column vector – thedilated Aug 15 '17 at 10:33
• what is $A$ defined to be? – mdave16 Aug 15 '17 at 15:26

For (a) note that for all $1\leq i\leq n$ we have $$(A-\lambda_1 I)v^{(i)}=(\lambda_i-\lambda_1)v^{(i)}$$ hence if $x=\sum_{i=1}^n\beta_iv^{(i)}$, then \begin{align} x^{(0)} &=(A-\lambda_1I)x\\ &=\sum_{i=1}^n\beta_i(A-\lambda_1I)v^{(i)}\\ &=\sum_{i=2}^n\beta_i(\lambda_i-\lambda_1)v^{(i)}\\ &=\sum_{i=2}^n\beta'_iv^{(i)} \end{align} where $\beta'_i=\beta_i(\lambda_i-\lambda_1)$ for $i\geq 2$.
Similarly, for (b) you have $$(A-\lambda_2 I)(A-\lambda_1 I)v^{(i)}=(\lambda_i-\lambda_2)(\lambda_i-\lambda_1)v^{(i)}$$ hence \begin{align} x^{(0)} &=(A-\lambda_2 I)(A-\lambda_1I)x\\ &=\sum_{i=1}^n\beta_i(A-\lambda_2 I)(A-\lambda_1I)v^{(i)}\\ &=\sum_{i=3}^n\beta_i(\lambda_i-\lambda_2)(\lambda_i-\lambda_1)v^{(i)}\\ &=\sum_{i=3}^n\beta'_iv^{(i)} \end{align} where $\beta'_i=\beta_i(\lambda_i-\lambda_2)(\lambda_i-\lambda_1)$ for $i\geq 3$.