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I have this confusion.

If $A$ is a matrix, then its eigenvectors are given by $(A-\lambda I)x = 0$. Since $x$ is not equal to zero, $A-\lambda I$ needs to be singular. I don't understand why $A-\lambda I$ needs to be singular. I thought $A-\lambda I$ needed to be zero, isn't it? Any pointers?

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There exists a case (in fact, many cases) where $A - \lambda I$ is not zero, $x$ is not zero, but $(A - \lambda I) x = 0$. –  Tunococ Aug 22 '12 at 23:43
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Let $(A - \lambda I)$ be an $n \times n$ matrix and $x$ be a non-zero $n \times 1$ vector. Let the lower case $c_i$ denote the columns of $A - \lambda I$. Now consider the following interpretation of $(A - \lambda I)x = \mathbf{0}$ as

$$ \pmatrix{\mid & \mid & & \mid \\ c_1 & c_2 & \dots & c_n \\ \mid & \mid & & \mid } \pmatrix{x_1 \\ x_2 \\ \vdots \\ x_n} = \mathbf{0}$$ which is equivalent to $$x_1 c_1 + x_2 c_2 + \dots + x_n c_n = \mathbf{0}.$$ If the $x_i$'s are non-zero, then either

  1. $c_1 = c_2 = \dots = c_n = \mathbf{0}$, and hence $A - \lambda I = \mathbf{0}$, or

  2. The set of vectors $\{ c_1, c_2, \dots, c_n \}$ is linearly dependent. In such case, we say that the kernel of $A - \lambda I$ is non-empty (non-trivial). For more information, read about nullspaces here and here.

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You say that we want $A-\lambda I$ to be zero. Watch out; what we actually want is to find $x$ such that $(A-\lambda I)x = 0$. It's a different thing.

So, we want to find non-zero vectors $x$ such that $(A-\lambda I)x=0$. This is a good old homogeneous system of linear equations. $x=0$ is always a solution of that, but it's sort of useless; the zero vector is always going to be an eigenvector so it's not very interesting. So, we remember that a homogeneous system has nontrivial (that is, $x\neq 0$) when its matrix is singular, or in other words, its determinant is $0$.

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This is another example of the fact that quantifiers are just as important as equations. It's not just "$(A - \lambda I) x = 0$", it's "there is some $x \ne 0$ such that $(A - \lambda I) x = 0$."

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