Answer to Question #5254 in Linear Algebra for Katelynn Bento




 Given Matrix A= |123|
 |023|
 |003|

Find the basis of Col A and the dim of Col A
Notice that the matrix A is upper-triangle, so its determinant is a  
product   
of diagonal elements:  
  
det(A) = 1*2*3 = 6.  
  
Moreover, det(A) is   
non-zero.  
Therefore the columns of A are linearly independent and   
therefore  
they constitutive a basis for R^3, so R^3 = Col(A).  
  
Hence   
any base of R^3 fits to be a base for Col(A),   
for instance the columns of   
A:  
  
(1,0,0), (2,2,0), (3,3,3)  
  
or   
  
(1,0,0), (0,1,0),   
(0,0,1)  
  
are the base.  
  
The dimension of Col(A) is the dimension of   
R^3 and therefore is equal  
to 3.
finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). **A Zero matrix**

∴ Let A be the symmetric and skew symmetric matrix.

⇒ A’=A (Symmetric)

⇒ A’=-A (Skew-Symmetric)

Considering the above two equations,

⇒ A=-A

⇒ 2A=0

⇒ A=0 (A Zero Matrix)

**Hence Zero matrix is both symmetric and skew symmetric matrix.** Simply using `df.mean()` will Do The Right Thing(tm) with respect to NaNs:



```
>>> df
                 A      B
DATE                     
2013-05-01  473077  71333
2013-05-02   35131  62441
2013-05-03     727  27381
2013-05-04     481   1206
2013-05-05     226   1733
2013-05-06     NaN   4064
2013-05-07     NaN  41151
2013-05-08     NaN   8144
2013-05-09     NaN     23
2013-05-10     NaN     10
>>> df.mean(axis=1)
DATE
2013-05-01    272205.0
2013-05-02     48786.0
2013-05-03     14054.0
2013-05-04       843.5
2013-05-05       979.5
2013-05-06      4064.0
2013-05-07     41151.0
2013-05-08      8144.0
2013-05-09        23.0
2013-05-10        10.0
dtype: float64

```

You can use `df[["A", "B"]].mean(axis=1)` if there are other columns to ignore. hello, 

this code checks to see if the diagonal elements of a given matrix a (assuming n x n) are larger in magnitude than the sum of the magnitude of the non-diagonal elements in its row.

line by line explanation:

the first line loops through all the rows of a.

 for i = 1:n

the second line is defining j = [1 2 3 ... n], so a vector containing 1..n,
 
 j = 1:n;

the third line is removing the diagonal element, aka, the ith row and this removes the ith column, so the i,i th element,
 
 j(i) = [];

the fourth line takes the absolute value of all the elements in the ith row of a, excluding the diagonal element: a(i,j) = a(ith row, elements 1,2,...,i-1,i+1,...,n) since we removed the ith element from j previously.
 
 b = abs(a(i,j));

the fifth line computes the absolute value of the diagonal element minus the sum of the absolute value of all the other elements in the row. if check is positive, we know the magnitude of the diagonal element was greater than the sum of all the magnitudes of the other elements in its row.
 
 check(i) = abs(a(i,i)) - sum(b);

the sixth line checks if it's less than zero, in which case the diagonal elements magnitude is not greater than or equal to the sum of the magnitudes of the other elements.
 
 if check(i) < 0

if this is the case, the matrix is not strictly diagonally dominant, which the seventh line prints.
 
 fprintf('the matrix is not strictly diagonally dominant at row %2i

',i)

the eighth and ninth lines end the cycle.
 
 end
 end

sorry if this was too much detail,

hope this helps! it's not clear what it is you want as an output. if it's a row vector, there's no concatenation involved, you just simply reshape your matrix:
a = [0 1 1 1;
 1 0 1 0;
 1 0 0 1]
 v = reshape(a, 1, []) %reshape a into one row. since matlab is column major, this is done column-wise solution: `abs(veca)=abs(vecb)=abs(vecc)=lamda`   

`veca*vecb=vecb*vecc*vecc*veca=0`   

`abs(veca+vecb+vecc)^2=(veca+vecb+vecc)*(veca+vecb+vecc)`   

`=abs(veca)^2+0+0+abs(vecb)^2+abs(vecc)^2`   

`=lambda^2+lambda^2+lambda^2`   

`=3lambda^2`   

`abs(veca+vecb+vecc)=sqrt3lambda`   

`costheta\_1=(veca*(veca+vecb+vecc))/(abs(veca)abs(veca+vecb+vecc))=lambda^2/(lambdasqrt3lambda)=1/sqrt3=theta\_1=cos^(-1)(1/sqrt3)`   

similarly,   

`theta\_1=theta\_2=theta\_3=cos^(-1)(1/sqrt3)` sin(theta) = d, 
cos(theta) = a, 
tan(theta) = h, 
csc(theta) = f, 
sec(theta) = c, 
cot(theta) = g. answer: `1/221`
solution: n/a given;y-axis, y = cos x and y = sin x, 0 ≤ x ≤ π/2⇒ sin x = cos xrequired area= (√2 − 1) sq.units solution: given that, `sinx+cosx=a`   
 on squaring both sides, we get   
 `" "(sinx+cosx)^(2)=(a)^(2)`   
 `rarr" "sin^(2)x+cos^(2)x+2sinxcosx=a^(2)`   
 `rarr" "sinx*cosx=(1)/(2)(a^(2)-1)`   
 `(i) sin^(6)x+cos^(6)x=(sin^(2)x)^(3)+(cos^(2)x)^(3)`   
 `" "=(sin^(2)x+cos^(2)x)(sin^(4)x-sin^(2)xcos^(2)x+cos^(4)x)` ltbrgt `" "=sin^(4)x+cos^(4)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=(sin^(2)x+cos^(2)x)^(2)-2sin^(2)xcos^(2)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=1-2*(1)/(4)(a^(2)-1)^(2)-(1)/(4)(a^(2)-1)^(2)=(1)/(4)[4-3(a^(2)-1)^(2)]`   
 `(ii) abs(sinx-cosx)=sqrt((sinx-cosx)^(2))`   
 `" "=sqrt(sin^(2)x+cos^(2)x-2sinxcosx)`   
 `" "=sqrt(1-2(1)/(2)(a^(2)-1))=sqrt(1-a^(2)+1)=sqrt(2-a^(2))`