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Using matrices as basis 
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#1
Jan213, 04:49 AM

P: 12

Hi. Define a linear mapping F: M2>M2 by F(X)=AXXA for a matrix A, and find a basis for the nullspace and the vectorspace(not sure if this is the term in english). Then I want to show that dim N(F)=dim V(F)=2 for all A, A≠λI, for some real λ. F(A)=F(E)=0, so A and E belongs to the nullspace. Then I define a basis for M2, as the 2x2matrices B=(B11, B12, B21, B22) which has a 1 at i,j and 0's elsewhere. Well, this is how Im supposed to do, but it confuses me.
How should I view the basismatrix? For example with linear independency. Lets say we define A to be the 2x2matrix with elements (a,b,c,d) and map them with F. We get 4 matrices F(Bij) and I want to sort out which ones are linearly independent, with the condition A≠λI. How do I show L.I for matrices? 


#2
Jan213, 07:13 AM

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PF Gold
P: 39,533

Of course, X is in the null space if and only If AX= XA. In other words, the space of all matrices that commute with A. I'm not sure what it would look like but you could "experiment" by looking at [itex]A= \begin{bmatrix}a & b \\ c & d \end{bmatrix}[/itex] and [itex]X= \begin{bmatrix}s & t \\ u & v\end{bmatrix}[/itex] and then you want to have [itex]AX= \begin{bmatrix}as+ bu & at+ bv \\ cs+ du & ct+ dv\end{bmatrix}= \begin{bmatrix}as+ ct & bs+ dt \\ au+ cv & bu+ dv\end{bmatrix}= XA[/itex] so that we must have as+ bu= as+ ct or just bu= ct, at+ bv= bs+ dt, cs+du= bu+ dv, and ct+dv= bu+ dv so that ct= bu. The one thing those have in common is ct= bu so that if one of t or u is 0, the other is also. If, in fact, both t and u are 0, the other equations become bv= bs so that v= s. That gives us [itex]\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}[/itex] as one basis vector and [itex]\begin{bmatrix}0 & 1 \\ c/b & 0\end{bmatrix}[/itex] as another. 


#3
Jan213, 11:59 PM

P: 12

I meant the column vectorspace, the range yes! I mixed up E and I, we use E for the identity matrix.
So, what we've done here is(if you could check if my line of thoughts is correct), made the transformation for an arbitrary matrix A and X. Examined the nullspace by setting AX=XA, and shown that the nullity of the transformation is 2. Furthermore, we've found 2 basis vectors with rank 2 of the transformation, so the dimN(F)=dimC(F)=2. I keep mixing things up, A is the matrix for the transformation right? Which has the basis vectors we've just shown? Ive never seen a matrix being a base before. When we have basis vectors, we just put them as columns for a transformation matrix, what do we do with basis matrices for a transformation matrix? This relates to the condition A≠λI how? Thx for answering HallsofIvy! 


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