Why is it that when multiplying a (1x3) by (3x1) matrix, you get a (1x1) matrix, but when multiplying a (3x1) matrix by a (1x3) matrix, you get a (3x3) matrix? Why is matrix multiplication defined this way?  Why can't a (1x3) by (3x1) yield a (3x3), or a (3x1) by (1x3) yield a (1x1)? I really would like to get to the root of this problem or 'axiomatization'. Thanks.
10744
If I multiply two numbers, say $3$ and $5$, I know it means add $3$ to itself $5$ times or add $5$ to itself $3$ times.    But If I multiply two matrices, what does it mean ?  I mean I can't think it in terms of repetitive addition.       What is the intuitive way of thinking about multiplication of matrices?
68120
I was following this explanation until the 5th step. The most important misunderstanding is if $1x/1x$ is $1$, and one can multiply a quotient by $1$ and not change its value. However from what I can see the $9x^2$ is being divided by $x^2$. How is that going to work? And why is that allowed?
45349
I want to know if the set $\{(1, 1, 1), (3, 2, 1), (1, 1, 0), (1, 0, 0)\}$ spans $\mathbb{R}^3$. I know that if it spans $\mathbb{R}^3$, then for any $x, y, z, \in \mathbb{R}$, there exist $c_1, c_2, c_3, c_4$ such that $(x, y, z) = c_1(1, 1, 1) + c_2(3, 2, 1) + c_3(1, 1, 0) + c_4(1, 0, 0)$.  I've looked around the internet, but all the answers I found involve setting up a matrix and finding the determinant, and I can't do that here because my matrix isn't square. What am I missing here?
17900
I'm trying to prove that for a certain matrix $A$, and its conjugate transpose $A^*$, we have $⟨Ax,y⟩=⟨x,A^*y⟩$, where $⟨⟩$ represent the inner product. So here it's simply the dot product in $R^n$.  Does anyone have an idea of how to prove this?  Thank you!
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I have just watched the first half of the 3rd lecture of Gilbert Strang on the open course ware with link:    It seems that with a matrix multiplication $AB=C$, that the entries as scalars, are formed from the dot product computations of the rows of $A$ with the columns of $B$. Visual interpretations from mechanics of overlpaing forces come to mind immediately because that is the source for the dot product (inner product).  I see the rows of $C$ as being the dot product of the rows of $B$, with the dot product of a particular row of $A$. Similar to the above and it is easy to see this from the individual entries in the matrix $C$ as to which elements change to give which dot products.  For understanding matrix multiplication there is the geometrical interpretation, that the matrix multiplication is a change in the reference system since matrix $B$ can be seen as a transormation operator for rotation, scalling, reflection and skew. It is easy to see this by constructing example $B$ matrices with these effects on $A$. This decomposition is a strong argument and is strongly convincing of its generality. This interpreation is strong but not smooth because I would find smoother an explanation which would be an interpretation begining from the dot product of vectors and using this to explain the process and the interpretation of the results (one which is a bit easier to see without many examples of the putting numbers in and seeing what comes out which students go through).  I can hope that sticking to dot products throughout the explanation and THEN seeing how these can be seen to produce scalings, rotations, and skewings would be better. But, after some simple graphical examples I saw this doesn't work as the order of the columns in matrix $B$ are important and don't show in the graphical representation.   The best explanation I can find is at . It is convincing but a bit disappointing (explains why this approach preserves the "composition of linear transformations"; thanks @Arturo Magidin). So the question is: Why does matrix multiplication happen as it does, and are there good practical examples to support it? Preferably not via rotations/scalings/skews (thanks @lhf).
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Let $r\in \mathbb{R}$ and let $0\neq s \in \mathbb{R}$. Define operations $\#$ and $\&amp;$ on $\mathbb{R}$ x $\mathbb{R}$ by $(a,b) \mathbin\#(c,d)=(a+c,b+d)$ and $(a,b) \mathbin\&amp;(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)$ as addition and multiplication, respectively. Do these operations form a field?  I have shown that every axiom of the field is satisfied except for the "existence of multiplicative inverses" axiom. This is where I need help. We either need to show that this axiom fails and so we don't have a field or that the axiom $aa^{-1}=1$ is satisfied for every $a\neq 0 \in \mathbb{R}$  The multiplicative identity in this problem is $(1,0)$.  Any solutions or hints are greatly appreciated.
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I was thinking about the problem that says:     If $A$ and $B$ are $3\times 3$ real matrices such that $\operatorname{rank}(AB)=1$, then $\operatorname{rank}(BA)$ can not be which of the following?   (a) $0$   (b) $1$   (c) $2$   (d) $3$.   My attempt: I have chosen suitable $3 \times 3 $ matrices for $A$ and $B$ keeping in mind that $\operatorname{rank}(AB)=1$. Say for example if I take $A$ and $B$ to be  $$A = \begin{pmatrix} 1 &amp;2  &amp;0 \\  0 &amp; 0 &amp;0 \\  0 &amp; 0 &amp;0  \end{pmatrix}$$  and  $$B = \begin{pmatrix} -2 &amp;1  &amp;0 \\  1 &amp; 0 &amp;0 \\  0 &amp; 0 &amp;0  \end{pmatrix}$$ respectively, then I see $\operatorname{rank}(AB) = \operatorname{rank}(BA) = 1$. So, option (b) can not be correct. Do I have to keep choosing the matrices and then observe which of the option holds good. Is this kind of approach right to tackle the problem? I am looking for a direct way (e.g. application to some theorem) which can give me the result. I have also noticed that $AB$ and $BA$ are similar matrices as we see that $A^{-1}(AB)A=BA$. Is this observation going to help me in any way? Thanks in advance for your time.
21095
I was hoping that someone could point me in the right direction on how I could start answering the following question:     Based on the fact that $\det(AB)=\det(A)\det(B)$ for all square $n \times n$ matrices $A,B$, explain in every detail why:   $$\det(A^{-1}) = 1/\det(A).$$
92343
As far as I know $(AB)^{-1}=B^{-1}A^{-1}$ and matrix multiplication is associative so:  $$(X′ X)^{−1}X′=(X^{-1})(X')^{-1}(X')=X^{-1}.$$  What am I doing wrong?
59108
Let $𝐺$ be the group generated by the matrices $𝐴=\begin{bmatrix}2&amp;0\\0&amp;1\end{bmatrix}$, $𝐵=\begin{bmatrix}1&amp;1\\0&amp;1\end{bmatrix}$, where the group operation is matrix multiplication.      Prove that $𝐺$ contains a subgroup $𝐻_𝑖$ for each $𝑖\in \mathbb{Z}$ such that   $$ \cdots \subset H_{-2}\subset H_{-1} \subset H_0 \subset H_1 \subset \cdots$$   So this is all that is given and i'm not sure where to start from this question. I know that $ABA^{-1}$ = $B^{2}$. From this we can replace A with $A^2$ and so on but i am not sure how i would prove this. Can someone help please?
58532
We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.  As background there are plenty of mathematical schemes that can be used to define these operators. One of these is   where $H_0(a,b) = b+1$   (successor op) $H_1(a,b) = a+b$   (addition op) $H_2(a,b) = ab $   (multiplication op) $H_3(a,b) = a^b$   (exponentiation op) $H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times ) etc.    Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$  Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?  Thanks.
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I think it wants me to show that there's no bijection between the two sets.  I first tried to show that there's simply no bijection, then I realised that it doesn't work.  If I'm to show there's no bijective morphism that carries multiplication in nonzero real numbers to addition in real numbers, how am I supposed to do it? I'm thinking about doing something with 0, since it's the element of the first set but not the second set.  Thanks!
36223
Is there any $2 \times 2$ matrix $A \neq I$ such that $A^3=I$?  In my opinion: no. Thank you very much.
115114
Question is: Find all values of a for which $S=\{v_1,v_2,v_3\}$  form a basis for ${\bf R}^3$, where $v_1=(a^2,0,1)$, $v_2=(0,a,2)$, and $v_3=(1,0,1)$.  My thoughts: I put it into a matrix to help visualize:  $$ S=\begin{bmatrix} a^2 &amp;0 &amp;1\\  0&amp; a &amp;0\\  1 &amp;2 &amp;1 \end{bmatrix}. $$  Well, I can see right away that if $a=\pm1$ then we have 2 equal vectors $\implies$ linearly dependent. I also see that if $a=0$ we have a row of zeros $\implies$ Not spanning ${\bf R}^3$. So, I know that a cannot equal $-1$, 0, or 1  My problems: 1. I'm not sure how to show that any other value will form a basis. 2. I don't know what how would I solve similar questions like this, where it's not as obvious.  Thank you!
78925
Let $A$ be the matrix described when $A \in M_n(\mathbb{R})$.  Let $a,b \in \mathbb{R}$.  Prove that:   Iv'e tried to factor out $a+b$, but got nowhere.  A hint or a solution would be very appreciated.  Thanks a lot!
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I know that there are many proofs regarding this. However, the book i'm using seems to suggest another way to do it without giving an answer. What i mean by the another way is some other proofs that do not use the fact that elementary row operation can be expressed by multiplying elementary matrices. The book says that the lemma need to be proved only when the size of identity matrix is 2 by 2. Does it have something to do with mathematical induction? That is, did the author want us to find out both induction hypothesis and induction step?
8333
How does one solve for the matrix $A$ in the system $$ Ax = b$$ when say, $A$ is a $3\times 3$ matrix and $x$ and $b$ are both $3 \times 1$. $x$ and $b$ are obviously known.  The obvious approach would be to Multiply $A$ and $x$ and compare coefficients. How does one solve this in a general setting.