If the kernel of a matrix is the $0$ vector, why is the basis of the kernel non existent?   If I have the matrix  $\begin{bmatrix}1 &amp; 2\\3 &amp; 4\end{bmatrix}$  The kernel is the $0$ vector  Why is the basis of the kernel non-existent? Shouldn't the basis of the kernel be $0$?
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According to C.H. Edwards' Advanced Calculus of Several Variables: The dimension of the subspace $V$ is defined to be the minimal number of vectors required to generate $V$ (pp. 4).  Then why does $\{\mathbf{0}\}$ have dimension zero instead of one? Shouldn't it be true that only the empty set has dimension zero?
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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 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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In linear algebra, the kernel of a matrix is its null space.  In machine learning and statistics, there are a bunch of matrices are called "kernel". For example,    I am totally confused. The second "kernal" concept looks very much like a projection to me, rather than a "null space". Why they are given the same name? Is there any connection between the two concepts?  Thank you!
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Let $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$ such that:    $$T\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix}0\\ y\end{pmatrix}$$  Show that there's no ordered-basis, $E$ such that:    $$[T]_E = \begin{pmatrix}    1 &amp; 2\\    2 &amp; 4\\   \end{pmatrix}$$  I didn't get how should I prove it. I tried some ways but end up with nothing. I guess I'm missing something.
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Does $\lambda \mathbf u = 0$  imply $\lambda = 0 $ or $\mathbf u= 0$ where $\mathbf u$ is in a vector space $V$ over a field $F$.   The case for the vector equalling zero is trivial due to the fact field axioms include a multiplicative inverse, so how could I show lambda is 0?
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A real $2 \times 2 $ matrix $M$ such that $$M^2 =  \begin{pmatrix} -1&amp;0 \\ 0&amp;-1-\epsilon  \\   \end{pmatrix}$$ (a) exists for all $\epsilon &gt; 0$. (b) does not exist for any $\epsilon &gt; 0$. (c) exists for some $\epsilon &gt; 0$. (d) None of the above is true.  Attempt: I couldn't think of any theory base to prove / disprove the existence of such a matrix. Could someone please give me a hint on how to go about this problem? Thank you very much for your help in this regard.
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(a) Given the column vector $a$ as an $n\times 1$ matrix. Write out its singular value decomposition, showing the matrices $U, \sum$, and $V$ explicitly.  (b) Given the row vector $a^T$ as an $1\times n$ matrix. Write out its singular value decomposition, showing the matrices $U, \sum$, and $V$ explicitly.  I currently get stucked on this question, since I couldn't find the eigenvalues of the $n\times n$ matrix $aa^T$ in part (a), and $a^Ta$ in part (b) to find the columns of $U$ and $V$, respectively. Can someone please help me on this problem?
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Let $E$ be $\mathbb C$-vector space of dimension $3$. Let $f$ be non zero endomorphism of $E$ such that $f^2=0$. show that there exists a basis $B=\{b_1,b_2,b_3\} $ of $E$ where the matrix of $f$ is given by :  $$\begin{array}{ccc} 0 &amp; 1 &amp; 0 \\  0 &amp; 0 &amp; 0 \\  0 &amp; 0 &amp; 0 \end{array}$$  My try : since $f$ is non zero, there exists $x\in E$ such that $f(x)\not =0$. Take $b_1=f(x)$ and $b_2=x$. Now $b_1,b_2$ can easily shown to be linearly independent. By the exchange lemma we can find $b_3$ in another generating system of $E$ to make $b_1,b_2,b_3$ into a basis of $E$. Now clearly $f(b_1)=f^2(x)=0$ and $f(b_2)=f(x)=b_1$. Now how to find  a $b_3$ such that $f(b_3)=0$ and $b_1,b_2,b_3$ is a basis of $E$? Thank you for your help.   Edit: For the linear independence of $b_1,b_2$, if $\alpha x+\beta f(x)=0$ then we apply $f$ to get $\alpha f(x)=0$ and since $f(x)\not =0$ then $\alpha=0$ it remains that $\beta f(x)=0$ and again this gives that $\beta=0$.
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It is quite simple to show that all finite-dimensional vector spaces with inner product have an orthogonal basis, with the standard definition of a basis from linear algebra. However, I am in trouble to find any reference about infinite-dimensional case.  Do all infinite-dimensional vector space with inner product have an orthogonal basis (again with the standard definition of a basis from linear algebra)?
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Suppose we have $n$ linearly independent vectors $\mathbf{v}_{1}\ldots\mathbf{v}_{n}$ in $\mathbb{R}^{n}$. I know that they do span $\mathbb{R}^{n}$, because we can easily specify a non-singular map which sends the $\mathbf{v}_{i}$s to the standard basis, and then to whichever vector in $\mathbb{R}^{n}$ we choose.    My question is: do we need all the machinery of linear maps, determinants, etc. or is there a proof which is closer to the definitions? Every time I start writing down a proof I end up wanting to say "and this set of equations can be solved uniquely because this matrix is non-singular". Is this necessary?
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A set of orthonormal vectors  is linearly independent.   Since I wouldn't memorize any such facts that is mentioned in my course book, what is the reason behind it and why are they linearly independent?
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I was trying to show that orthogonal matrices have eigenvalues $1$ or $-1$.  Let $u$ be an eigenvector of $A$ (orthogonal) corresponding to eigenvalue $\lambda$. Since orthogonal matrices preserve length, $ \|Au\|=|\lambda|\cdot\|u\|=\|u\|$. Since $\|u\|\ne0$, $|\lambda|=1$.  Now I am stuck to show that lambda is only a real number. Can any one help with this?
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If A is invertible, prove that $\lambda \neq 0$, and $\vec{v}$ is also an eigenvector for $A^{-1}$, what is the corresponding eigenvalue?  I don't really know where to start with this one. I know that $p(0)=det(0*I_{n}-A)=det(-A)=(-1)^{n}*det(A)$, thus if both $p(0)$ and $det(0) = 0$ then $0$ is an eigenvalue of A and A is not invertible. If neither are $0$, then $0$ is not an eigenvalue of A and thus A is invertible. I'm unsure of how to use this information to prove $\vec{v}$ is also an eigenvector for $A^{-1}$ and how to find a corresponding eigenvalue.
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The exponential kernel is defined by:  $$k(x,z) = e^{-\alpha\|x-z\|}$$ where $\alpha&gt;0$, $x,z\in \Bbb{R}^d$, $\|x\|$ is the 2-norm.  The kernel matrix is defined by $K_{ij} = k(x_i,x_j)$, $i,j\in[1\ldots n]$.  How to prove that $K$ is a positive (semi-positive) definite matrix?
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what do it mean by "A real nxn matrix always has a real eigenvalue when n is odd?" Shouldnt it be that regardless of odd or even there will always have a real root?  hope to hear from you guys soon:)
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Let $$A =\begin{bmatrix} 1&amp;3&amp;-6&amp;2\\5&amp;12&amp;-24&amp;7\\9&amp;15&amp;-30&amp;6\end{bmatrix}$$  $$RREF(A) =\begin{bmatrix} 1&amp;0&amp;0&amp;-1\\0&amp;1&amp;-2&amp;-1\\0&amp;0&amp;0&amp;0\end{bmatrix}$$ $$REF(A) =\begin{bmatrix} 1&amp;3&amp;-6&amp;2\\0&amp;1&amp;-2&amp;1\\0&amp;0&amp;0&amp;0\\\end{bmatrix}$$  Now, provided this information, it can be found that the basis for the nullspace of $A$ using the $RREF(A)$ is: $$\begin{pmatrix} \begin{bmatrix} 0\\2\\1\\0 \end{bmatrix}, \begin{bmatrix} 1\\-1\\0\\1 \end{bmatrix}\end{pmatrix}$$  Using the $REF(A)$ the "basis" for the nullspace is:  $$\begin{pmatrix} \begin{bmatrix} 6\\2\\1\\0 \end{bmatrix}, \begin{bmatrix} -2\\-1\\0\\1 \end{bmatrix}\end{pmatrix}$$  Is the "basis" that I provided for the $REF(A)$ actually a basis? How does the  $REF$ and $RREF$ of a matrix differ?