```Question 62805
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AT THE OUTSET YOU HAVE DONE LET ME SAY THAT YOU HAVE DONE PERFECTLY WELL
SEE MY COMMENTS BELOW.I AM GIVING ADDITIONAL SUGGESTIONS FOR IMPROVEMENT
IN BRACKETS WHICH YOU MAY OMIT IN FIRST READING.
Hi,my name is Natalia. I solved two problems, but I'm not sure that I did it right. I would really appreciate if you can check them.
Here are two given problem:
Consider a vector space V=Mat 3x3 of all 3x3 matrices
Q.: determine the dimension of V,find and verify a basis for it.
here is what i've got so far:
if the matrix is
/a b c/
/d e f/
/g h i/OK....[IF YOU ARE DEALING WITH SMALLER MATRICES..YOU BETTER GET
ACCUSTOMED TO PUT IT AS
A1,A2,A3
B1,B2,B3
C1,C2,C3....WHERE COLUMNS ARE DESIGNATED BY 1,2,3 AND ROWS BY A,B,C, ETC
OR BETTER STILL
A11,A12,A13
A21,A22,A23
A31,A32,A33..WHERE ROWS ARE DENOTED BY I DIGIT AND COLUMNS BY SECOND DIGIT IN THE SUFFIX..KEEPING A SAME FOR THIS MATRIX WHICH WE CAN CALL AS MATRIX A. THEN WE CAN CALL ANOTHER MATRIX B BY FOLLOWING SAME TERMINOLOGY.B11,B12..ETC]

then the basis for it is....
-----------------------------------------------------
HERE I THINK WE BETTER REFLECT CLEARLY ABOUT 'BASIS'
LET US DISTINGUISH 3 THINGS FIRST.
1.SET OR OUR UNIVERSE:- IN OUR SET OR UNIVERSE UNDER CONSIDERATION ,
THERE COULD BE 1 OR 2 OR 3...
OR N MATRICES IF IT IS A FINITE SET OR INFINITE MATRICES IF IT IS
INFINITE SET AS IN YOUR
PROBLEM NOW..THERE ARE INFINITE NUMBER OF 3x3 MATRICE.IF IT IS
FINITE AND THERE ARE N MATRICES WE SAY OUR UNIVERSE IS A SET OF N GIVEN MATRICES.
2.DIMENSION:- THIS IS THE MINIMUM NUMBER OF LINEARLY INDEPENDENT MATRICES
SAY R MATRICES NEEDED TO BE TAKEN FROM THE ABOVE UNIVERSE TO EXPRESS
EVERY MATRIX IN THE UNIVERSE AS A LINEAR COMBINATION OF THESE R
MATRICES.R<=N ALWAYS.
3.BASIS:-YOU WILL FIND THAT THE ABOVE R MATRICES ARE NOT UNIQUE
IN GENERAL, UNLESS R=N, IN THE
SENSE THAT ANY SET OF R LINEARLY INDEPENDENT MATRICES FROM THE UNIVERSE
COULD BE PICKED UP TO EXPRESS EVERY MATRIX IN THE UNIVERSE AS A LINEAR
COMBINATION OF THOSE R MATRICES.ANY SUCH SET OF R MATRICES , WE CALL
A BASIS FOR THE UNIVERSE.THERE COULD BE BASIS B1,BASIS B2 ETC..
BUT ALL OF THEM WILL HAVE R AND ONLY R NUMBER OF LINEARLY
INDEPENDENT MATRICES.OK.?
ELEMENTARY BASIS:- THE EASIEST BASIS TO CONCEIVE IS (1,0,0,0....ETC),
(0,1,0,0...ETC),(0,0,1,0...ETC)....ETC...DEPENDING ON THE DIMENSION.
FOR 3D , WE WILL HAVE 3 , FOR 4D...4 ETC..
------------------------------------------------
NOW WE ARE TRYING TO PUT UP A BASIS FOR A UNIVERSE OF INFINITE SET
OF MATRICES.BUT FIRST LET US BE CLEAR ABOUT ITS DIMENSION.SINCE IT
IS A UNIVERSE OF 3X3 MATRICES , EACH OF WHICH CANHAVE 3*3=9
ELEMENTS,ALL INDEPENDENT IN THE SENSE THAT THEY CAN TAKE ANY
VALUE FROM THE REAL FIELD, THE DIMENSION WE CAN GUESS TO BE 3*3=9.[IN
NERAL FOE A NXN MATRIX WE HAVE DIMENSION OF N^2
NEXT IS TO PUT UP A....STANDARD E BASIS.
THIS IS EASILY DONE BY 9 MATRICES M1,M2,M3...M9 AS FOLLOWS.
M1=
1,0,0
0,0,0
0,0,0
-----------------
M2=
0,1,0
0,0,0
0,0,0
-------------
M3=
0,0,1
0,0,0
0,0,0
----------------
M4=
0,0,0
1,0,0
0,0,0
----------------ETC..AS IN SUCH A CASE ANY MATRIX 'A' CAN BE GOT BY
A11*M1+A12*M2+....ETC....A33*M9....WHERE A IS MATRIX EXPLAINED IN
THE BEGINING AS YOU HAVE DONE CORRECTLY
/1 0 0/ /0 1 0/ /0 0 1/ /0 0 0/ /0 0 0/ /0 0 0/ /0 0 0/ /0 0 0/ /0 0 0/
/0 0 0/,/0 0 0/,/0 0 0/,/1 0 0/,/0 1 0/,/0 0 1/,/0 0 0/,/0 0 0/,/0 0 0/
/0 0 0/ /0 0 0/ /0 0 0/ /0 0 0/ /0 0 0/ /0 0 0/ /1 0 0/ /0 1 0/ /0 0 1/
So, i have 9 vectors v1, v2,...,v9 that are lineraly independent and span
therefore, dimV=9
PERFECT!!

if the matrix is symmetric dimV=6
INDEPENDENT ELEMENTS WE HAVE HERE ARE 1+2+3....N IN I,II,....NTH.ROW
=N(N+1)/2 IN GENERAL ...HERE 3*4/2=6
EXCELLENT
and if the matrix is skew-symmetric dimV=9 Am I correct?
NO !DEAR!YOU MADE A SLIP HERE.
FROM THE ABOVE ANSWER,WE HAVE TO SUBTRACT N DIAGONAL ELEMENTS AS
THEY ARE ALL ZEROS.SO WE GET N(N+1)/2 - N = N(N-1)/2 ..
THAT IS 3*2/2=3 IS THE ANSWER.
And the second problem: I SHALL ANSWER LATER.LET ME POST THIS
FIRST FOR YOU.
==============================================
A have a matrix A=
/1 2 3 /
/2 5 7 /
/3 7 10/OK
that lookes like A=
/1 0 1/
/0 1 1/
/0 0 0/ when row-reduced
VERY GOOD
I need to express a vector x=
/2 /
/-1/
/7 /OK
as a sum of a vector from the row space and a vector from nullspace
For nullspace I got (let's call it v so it will be easier to write an expression then)
/-1/
/-1/
/1 /
EXCELLENT
for row space (u and w respectively)
/1/ /0/
/0/ /1/
/1/ /1/
VERY GOOD
that gives me 3-D subspace
and the vector expression is:
x=2*v + 4*u + 1*w
EXCELLENT !!!KEEP IT UP.YOU ARE REALLY GOOD!ONLY YOU NEED A
LITTLE SELF CONFIDENCE. BELIEVE IN YOUR ABILITY!GREAT JOB!!
did I solve it correct?
Thank you very much!```