[Math] Why is $\lambda I – A$ singular if $\lambda$ is an eigenvalue of $A$

eigenvalues-eigenvectorslinear algebra

Taken from Wikipedia, this is what's got me confused:

I'm not understanding how v being non-zero implies that its multiplier $(\lambda I – A)$ must be singular.

Also, I didn't quite get why the roots of the determinant of $(\lambda I – A)$ are the eigenvalues for $A$.

I looked at this post for some hints and I did some searching, but didn't really find a good answer.

As a side note, I'm going to go buy a linear algebra textbook today as soon as the bookstore opens. I realize this is fundamental stuff that I should already know.

Thanks in advance.

Best Answer

I'd like to mention that there are many characterisations of the singularity of a matrix (see for instance here).

If $\lambda I - A$ is singular, this means that there is non-zero vector $v$ such that $(\lambda I - A)v=0$, which can be rewritten as $Av = \lambda v$. Hence $v$ is an eigenvector.

Also, $\lambda I - A$ being singular is equivalent with $\det(\lambda I - A)=0$. Combining this with the reasoning hereabove shows that roots of $\det(\lambda I - A)=0$ are eigenvalues.

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[Math] Why is the eigenvalue equation $ Ax = \lambda x $ nonlinear

It sounds like you're interested in why "equation problems" $Ax=b$ are called linear while "eigenvalue problems" $Ax=\lambda\cdot x$ are called nonlinear. There are likely many reasons to use this language, but here's one that invokes the notion of a linear combinations.

Suppose that $x_1$ and $x_2$ are two solutions to "equation problems" $Ax_1=b_1$ and $Ax_2=b_2$. Consider an arbitrary linear combination $x$ of $x_1$ and $x_2$, so $x=c_1\cdot x_1+c_2\cdot x_2$. If we define $b=c_1\cdot b_1+c_2\cdot b_2$, then $$ Ax = A(c_1\cdot x_1+c_2\cdot x_2) = c_1\cdot Ax_1+c_2\cdot Ax_2 = c_1\cdot b_1+c_2\cdot b_2 = b $$ This illustrates that "linear combinations of solutions to equation problems are solutions to equation problems."

Now, suppose that $x_1$ and $x_2$ are two solutions to "eigenvalue problems" $Ax_1=\lambda_1\cdot x_1$ and $Ax_2=\lambda_2\cdot x_2$. Again consider an arbitrary linear combination $x=c_1\cdot x_1+c_2\cdot x_2$. This linear combination $x$ solves an eigenvalue problem if $Ax=\lambda\cdot x$ for some $\lambda$. However, we have $$ Ax = A(c_1\cdot x_1+c_2\cdot x_2) = c_1\cdot Ax_1+c_2\cdot Ax_2 = c_1\cdot\lambda_1\cdot x_1+c_2\cdot\lambda_2\cdot x_2 \overset{?}{=} \lambda\cdot(c_1\cdot x_1+c_2\cdot x_2)=\lambda\cdot x $$ The $?$ in this equation indicates that a suitable $\lambda$ might not exist. Indeed, it isn't difficult to find exmamples where such a $\lambda$ does not exist. For instance, consider the data \begin{align*} A &= \left[\begin{array}{rr} 23 & 32 \\ -16 & -25 \end{array}\right] & \lambda_1 &= 7 & x_1 &= \left[\begin{array}{r} 2 \\ -1 \end{array}\right] & \lambda_2 &= -9 & x_2 &= \left[\begin{array}{r} 1 \\ -1 \end{array}\right] \end{align*} Note that $Ax_1=\lambda_1\cdot x_1$ and $Ax_2=\lambda_2\cdot x_2$. However, for $x=x_1+x_2$, we have $$ \overset{A}{\left[\begin{array}{rr} 23 & 32 \\ -16 & -25 \end{array}\right]}\overset{x}{\left[\begin{array}{r} 3 \\ -2 \end{array}\right]} = \left[\begin{array}{r} 5 \\ 2 \end{array}\right] \neq \lambda\cdot x $$

Side Note. As mentioned in the comments, there are other reasons to call the eigenvalue problem $Ax=\lambda\cdot x$ nonlinear. For example, if one uses the characteristic polynomial $\chi_A(t)=\det(t\cdot I_n-A)$ to solve for the eigenvalues of $A$, then one ends up factoring an $n$th degree polynomial, which is a nonlinear problem.

It's also worth noting that taking any linear combination of two eigenvectors $x_1$ and $x_2$ corresponding to the same eigenvalue $\lambda$ does yield a solution to an eigenvalue problem. This is precisely the statement that eigenvectors corresponding to an eigenvalue $\lambda$ are organized into the eigenspace $E_\lambda=\operatorname{Null}(\lambda\cdot I_n-A)$.

[Math] Proof that $\text{det}(AB) = \text{det}(A)\text{det}(B)$ without explicit expression for $\text{det}$

Disclaimer: I am cheating a little bit here because

In some sense, I just proved that two different definitions of the determinant are equal (namely the one that the OP gave and the one using multilinear forms).
One would still have to prove that the determinant as defined by the OP is continuous (which must be true). Perhaps one can use that the characteristic polynomial varies continuously as you move continuously through $\text{End}(V)$.

Let $\Lambda^n(V)$ denote the space of alternating $n$-multilinear transformations. For each $T\in\text{End}(V)$ there is a pull-back map $T^*:\Lambda^n(V)\rightarrow\Lambda^n(V)$ defined by $$ T^*(\omega)(v_1,\ldots,v_n)=\omega(T(v_1),\ldots,T(v_n)) ,$$ for each $\omega\in\Lambda^n(V)$ and $v_1,\ldots,v_n\in V$. Since $\Lambda^n(V)$ is $1$-dimensional, one can define $$\widetilde{\det}:\text{End}(V)\rightarrow \mathbb{C}$$ by setting $$ T^*(\omega)=\widetilde{\det}(T)\omega,$$ for each $\omega\in\Lambda^n(V)$. Then, by @hm2020's answer, it is clear that $\widetilde{\det}(AB)=\widetilde{\det}(A)\widetilde{\det}(B)$ (it follows from the definition at once). Now, for each diagonalizable $A\in\text{End}(V)$, take a basis of eigenvectors $\{e_i\}_{i=1}^n$ with eigenvalues $\{\lambda_i\}_{i=1}^n$ and $n$ vectors $v_k=\displaystyle\sum_{i=1}^n v^i_ke_i\in V$, $k\in\{1,\ldots,n\}$. It follows that \begin{equation*}\begin{split}T^*(\omega)(v_1,\ldots,v_n)&=\omega(T(v_1),\ldots,T(v_n))\\&=\omega(\sum v^i_1\lambda_ie_i,\ldots,\sum v^i_n\lambda_ie_i)\\ &=\left(\displaystyle\prod_{i=1}^n\lambda_i\right)(\omega(v_1,\ldots,v_n))\\ &=\det(T)\omega(v_1,\ldots,v_n), \end{split}\end{equation*} where we used that $\omega$ is alternating. It follows that $\det=\widetilde{\det}$ on the set of diagonalizable matrices, which is dense in $\text{End}(V)$. Since both of them agree on a dense set and are continuous (and the codomain is Hausdorff), we have $\det(T)=\widetilde{\det}(T)$ for every $T\in\text{End}(V)$. In particular, for every $A,B\in\text{End}(V)$, there holds $$ \det(AB)=\widetilde{\det}(AB)=\widetilde{\det}(A)\widetilde{\det}(B)=\det(A)\det(B).$$

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[Math] Why is the eigenvalue equation $ Ax = \lambda x $ nonlinear

[Math] Proof that $\text{det}(AB) = \text{det}(A)\text{det}(B)$ without explicit expression for $\text{det}$

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