When we want to get the eigenvalues of a square matrix $A$ we calculate $$ det(A - \lambda I) = 0$$ What I don't understand is that what we do here in basic terms. Like, for a simple, square matrix M, if determinant is 0 then the vectors are linearly dependent and the equation system described by M does not have a unique solution (it either has . But what about $A - \lambda I$, how can I describe what happens in $det(A - \lambda I) = 0$ in basic terms? I am trying to understand how the eigenvalues relate to this determinant. I know that eigenvectors S and eigenvalues $\lambda$ are related by $AS=\lambda S$. Please don't overwhelm me with proofs and math-heavy reasoning, I have a very basic understanding of linear algebra.
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I understand that the eigenvectors of a linear transformation are the vectors that, when transformed, are simply scaled by a factor of the corresponding eigenvalue. Therefore $$ A\vec{x}=\lambda\vec{x}. $$ We can replace the scaling of the vector $\vec{x}$ by $\lambda$ with its equivalent linear transformation, $\lambda I$. Then we can rearrange this equation as follows: $$ A\vec{x}-(\lambda I)\vec{x}=\begin{bmatrix}0\\0\end{bmatrix}\\  (A-\lambda I)\vec{x}=\begin{bmatrix}0\\0\end{bmatrix}\\  $$ For this to be true either $\vec{x}$ must be a zero vector or the transformation $A-\lambda I$ must have $\vec{x}$ in its null-space. The course I'm following says that we can solve for $\lambda$ by finding when the determinant of $A-\lambda I$ is zero, but why does that guarantee that $\vec{x}$ will become the zero vector when transformed by $A-\lambda I$? I understand that for $\vec{x}$ to be an eigenvector this transformation must have a determinant of zero, but I don't see how the inverse can be true; having a determinant of zero doesn't gaurantee that $\vec{x}$ will be transformed into the zero vector. For example, let $$ (A-\lambda I) = \begin{bmatrix}1&amp;2\\0&amp;0\end{bmatrix}, \vec{x}=\begin{bmatrix}1\\1\end{bmatrix} $$ Then, although $\det\begin{bmatrix}1&amp;2\\0&amp;0\end{bmatrix} = 0$, it does not gaurantee $(A-\lambda I)\vec{x}$ to be the zero vector. $$ \begin{bmatrix}1&amp;2\\0&amp;0\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\0\end{bmatrix}  $$ We can see that the transformation gives us $\begin{bmatrix}3\\0\end{bmatrix}$ and not the zero vector. Why does finding when the determinant of this transformation is equal to zero help us find the eigenvectors? Am I missing something?
109695
If $A^2=-I$ , where $A$ is a square matrix of order $n$ and which contains real entries only and $I$ is identity matrix. Then how can we prove that $\det(A)=1$?.  I could prove that $n$ should be an even integer. But could not proceed to prove that $\det(A)$ can take only $1$, finding out few matrices which satisfies such properies (of small order) also verifies the given statement that the determinant is only $1$ and not $-1$.  Can anyone help with a hint ?
80914
The question is: Prove that if $\lambda$ is an eigenvalue of a matrix A with corresponding eigenvector x, then $\lambda^2$ is an eigenvalue of $A^2$ with corresponding eigenvector x.  I assume I need to start with the equation $Ax=\lambda x$ and end up with $A^2 x=\lambda^2 x$ but between those I am kind of lost. I have manipulated the equations several different ways and just can't seem to end up where I need to be. Help would be greatly appreciated as I believe this will be on a test tomorrow.
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Assume that $A$ is an $n\times n$  real matrix, i.e. $$A^T=-A.$$  Since $\det(A-\lambda I)=\det(A^T-\lambda I)$, $A$ and $A^T$ have the same eigenvalues. On the other hand, $A^T$ and $-A$ also have the same eigenvalues. Thus if $\lambda$ is an eigenvalue of $A$, so is $-\lambda$. If $n$ is odd, $\lambda = 0 $ is an eigenvalue.  A curious search in Google returns  the nonzero eigenvalues of $A$ are all pure imaginary and thus are of the form $iλ_1, −iλ_1, iλ_2, −iλ_2,$ … where each of the $λ_k$ are real.   Here is my question:     How can I prove the fact that "the nonzero eigenvalues of $A$ are all pure imaginary"?
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The question is motivated from the following problem:     Let $I\neq A\neq -I$, where $I$ is the identity matrix and $A$ is a real $2\times 2$ matrix. If $A=A^{-1}$, then the trace of $A$ is   $$ (A) 2 \quad(B)1 \quad(C)0 \quad (D)-1 \quad (E)-2$$   Since $A=A^{-1}$, $A^2=I$. If the converse of  is true, then $\lambda^2=1$ and thus $\lambda=\pm1$. And then $\rm{trace}(A)=1+(-1)=0$.   Here are my questions:   Is  $C$ the answer to the quoted problem?  Is the converse of Cayley-Hamilton Theorem, i.e.,"for the square real matrix $A$, if $p(A)=0$, then $p(\lambda)$ is the characteristic polynomial of the matrix $A$" true? If it is not, then what's the right method to solve the problem above?
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I'm trying to prove that an involutory matrix (a matrix where $A=A^{-1}$) has only eigenvalues $\pm 1$.  I've been able to prove that $det(A) = \pm 1$, but that only shows that the product of the eigenvalues is equal to $\pm 1$, not the eigenvalues themselves.  Does anybody have an idea for how the proof might go?  Thanks.
58764
Let $A$ be a nonsingular $n \times n$ matrix, $\|\cdot\|$ be any natural norm, and $K_{p}(A)=$ $\|A\|_{p}\left\|A^{-1}\right\|_{p} .$ Let $\lambda_{1}$ be the smallest and $\lambda_{n}$ be the largest eigenvalues of the matrix $A^{t} A$ (b) Show that $K_{2}(A)=\sqrt{\frac{\lambda_{n}}{\lambda_{1}}}$. (Hint: One should use the $\|A\|=\sqrt{\rho\left(A^{t} A\right)}=\sqrt{\rho\left(A A^{t}\right)}$ relation) For part $(b)$, it seems minimal eigenvalue of matrix $A$ will be the largest eigenvalue of the inverse of matrix $A$, so we have to use this idea but could not understand formally why square roots are needed for $K_{2}(A)$ and how spectral radius is necessary for this part?!
16411
Suppose that $A \in \mathbb{R}^{n\times n }$, and $x,y\in \mathbb{R}^n$.    Show that there exist numbers $a,b$ so that $\det{(A + sxy^*)}=a+bs$   Show that if $\det{(A)}\neq 0$ then $a = \det{(A)}$ and $b = \det{(A)} \cdot y^*A^{-1}x$.   Is it true that if $\det{(A)}=0$ then $a=0$?    My answer so far: The third question seems trivial and silly, and I don't understand why it's on there. I have solved the first one as follows: take $\beta$ to be any basis whose first element is $x$. Then with respect to basis $\beta$, $A+ sxy^*$ has the form   $$\left(  \begin{matrix} \alpha_{11} + sy_1 &amp; \alpha_{12} + sy_2  &amp; \dotsb &amp; \alpha_{1n}+sy_n \\ \alpha_{21}&amp;\dotsb&amp;&amp;\alpha_{2n} \\ \vdots &amp; \ddots &amp;&amp;\vdots\\ \alpha_{n1} &amp; \dotsb &amp;&amp;\alpha_{nn} \end{matrix} \right),$$  from which it's clear that the determinant is a polynomial in $s$ of degree $\leq 1$.    I am less clear on the second question.   For a limited case, note that if $A$ is diagonalizable and $x$ is an eigenvector of $A$ with eigenvalue $\lambda_x \neq 0$, then adding $sxy^*$ to $A$ has the effect of preserving all the eigenvectors, bumping up $\lambda_x$ to $\lambda_x + sy^*x$, and since the determinant is the product of all the eigenvalues, the product of $\lambda_1 \lambda_2  \dotsb (\lambda_x + sy^*x)\dotsb \lambda_n $ is   $$\det{(A)} + \det{(A)}\frac{sy^*x}{\lambda_x}=\det{(A)}\cdot sy^*(\frac{x}{\lambda_x}) = \det{(A)}\cdot s y^*(A^{-1}x).$$  Note that $y^*A^{-1}x$ is $\alpha_1 y_1 + \dotsb + \alpha_n y_n$, for $\alpha_i$ such that $x = \alpha_1 A_1 + \dotsb + \alpha_n A_n$, where $A_i$ are the columns of $A$.  But this is all I've got. Can someone give me a hint? I'm not looking for you to feed me the answer. Thanks as always
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We have to find $\det(A^{2}+A^{T})$. It is given that eigenvalues of $A$ are $1,2,3$.  My attempt: Since the question implicitly states that the answer would be same for all $A$ with eigenvalues $1,2,3$ we find that the answer is $(1+1)*(2+4)*(3+9)$ by taking $A$ as a diagonal matrix :P  I am unable to prove it for general $A$. Any help will be appreciated.
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Does anyone know how this equation came.I was studying about Eigenvalues and a doubt that arised was how could the equation $(A- \lambda)V = 0$ become $|A- \lambda I| = 0$.
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So the question asks: Let A be a $2\times 2$ matrix, and suppose that there exist scalars $λ_1,λ_2$ in$\mathbb{R}$ and nonzero vectors $v_1,v_2$ in $\mathbb{R}$ such that $Av_1=λ_1v_1, Av_2=λ_2v_2$, and $λ_1$ does not equal to $λ_2$.Prove that the vectors $v_1,v_2$ form a basis of $R^2$.   So so far I got:   As the eigenvector $v_1,v_2$ are corresponding to two different eigen values$λ_1,λ_2$ respectfully.   Therefore the eigenvectors $v_1,v_2$ are linearly independent.   And since $v_1,v_2$ spans A in $\mathbb{R}$.   Hence,  $v_1,v_2$ forms a basis of $\mathbb{R}$.   Does this proof look right? Do I need to write a linear combination of $v_1,v_2$ to show they span A?
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Let 'A' be a $2 \times 3$ matrix where as B be a $3 \times 2$ matrix if $\det(AB) = 4$ the find value of the $\det(BA)$  My attempt: I took A =  $$         \begin{bmatrix}         2 &amp; 0  &amp;0\\         0 &amp; 0  &amp;2\\         \end{bmatrix} $$  B= $$         \begin{bmatrix}         1 &amp; 0  \\         0 &amp; 0  \\         0 &amp; 1  \\         \end{bmatrix} $$  It satisfies given condition and I get $\det(BA)=0$   But I have not proved it  How do I prove that it is always zero (background)I am 12th grader and I know about adjoint,inverse,determinant,rank of a matrix and the other basics.  However I do NOT know about eigenvalues and eigenvectors.
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If $\lambda$ is a real eigenvalue of a real matrix $M$, does there necessarily exist a real eigenvector of $M$ corresponding to $\lambda$?  Edit: Never mind. I figured it out.. If $\lambda$ is a real eigenvalue, then $\det(\lambda I-M)=0$, which means there exists a real vector $x$ such that $Mx=\lambda x$. Right?
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If $A$ is idempotent (defined by $AA=A$) then the eigenvalues of $A$ are $0$ or $1$.   Proof: $$Ax = \lambda x$$ $$\Rightarrow Ax = AAx = \lambda Ax = \lambda^{2}x.$$  Then in the lecture notes, it says that the above implies that $\lambda^{2} = \lambda$ which implies $\lambda = 0$ or $\lambda = 1$.   I don't quite follow how it is deduced that $\lambda^{2} = \lambda$ from the above calculations.
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If I have a $2\times 2$ or $3\times 3$ matrix, how should I go about computing the eigenvalues and eigenvectors of the matrix?  NB: I am making this question to provide a unified answer to questions about eigenvalues of small matrices so that all of the specific examples that come up can be marked as duplicates of this post. See .
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This is a statement in linear algebra that I can't seem to understand the proof behind.  For a square matrix $A$, why is: $$\mathrm{adj}(A)A = \det(A) \cdot I$$  Any explanation would be greatly appreciated.
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While I was solving problem to determine if given set of vectors are linearly independent or not, the solution given was that the coefficient matrix A of the homogeneous linear system of those vectors   k1a1 + k2b1 + k3c1 = 0  k1a2 + k2b2 + k3c2 = 0  k1a3 + k2b3 + k3c3 = 0  was a 3x3 square matrix, and determinant(A) = 0  Thus it has a non trivial solution and hence the vectors are linearly dependent.  I am unaware about how if A is nxn square matrix and if determinant(A) = 0, then the system is having a non trivial solution.