Let A be a matrix which is both symmetric and skew symmetric.
so AT=A and AT= -A
so A =- A
that implies 2A =zero matrix
that implies A is a zero matrix
the matrix whose entries are all 0
A foreign can have null values and it can have duplicate values.A foreign can have null values and it can have duplicate values.A foreign can have null values and it can have duplicate values.A foreign can have null values and it can have duplicate values.A foreign can have null values and it can have duplicate values.A foreign can have null values and it can have duplicate values.A foreign can have null values and it can have duplicate values.A foreign can have null values and it can have duplicate values.A foreign can have null values and it can have duplicate values.A foreign can have null values and it can have duplicate values.A foreign can have null values and it can have duplicate values.
The null space describes what gets sent to 0 during the transformation. Also known as the kernel of the transformation. That is, for a linear transformation T, the null space is the set of all x such that T(x) = 0.
No. It can be infinite, finite or null. The set of odd integers is infinite, the set of even integers is infinite. Their intersection is void, or the null set.
A set with no numbers in it; also called an empty set.
yes, it is both symmetric as well as skew symmetric
I could be wrong but I do not believe that it is possible other than for the null matrix.
The null matrix is also called the zero matrix. It is a matrix with 0 in all its entries.
Yes.
Yes.
the matrix whose entries are all 0
the matrix whose entries are all 0
If x is a null matrix then Ax = Bx for any matrices A and B including when A not equal to B. So the proposition in the question is false and therefore cannot be proven.
there is none you weasel. the only good matrix is revolutions. :)
nullity of A is the dimension of null space of A.
means whether the matrix is same or not program for symmetric matrix : include<stdio.h> #include<conio.h> main() { int a[10][10],at[10][10],k,i,j,m,n; clrscr(); printf("enter the order of matrix"); scanf("%d %d",&m,&n); printf("enter the matrix"); for(i=0;i<m;i++) { for(j=0;j<n;j++) scanf("%d",&a[i][j]); } for(i=0;i<m;i++) { for(j=0;j<n;j++) at[i][j]=a[j][i]; } for(i=0;i<m;i++) { for(j=0;j<n;j++) { if(at[i][j]!=a[i][j]) k=1; } } if(k==1) printf("not symmetric"); else printf("symmetric"); getch(); }
No. But then can you prove that you do?