QR Factorization Suppose A is an m Times n matrix with m Greaterthanorequalto n.

Question

QR Factorization Suppose A is an m Times n matrix with m Greaterthanorequalto n. Suppose the first n - 1 columns of A are linearly independent, but the last column is a linear combination of the previous columns (i.e. A has rank n - 1). Can we still construct an m Times n orthogonal matrix Q using Gram-Schmidt? Yes, the final column will be a linear combination of the first n - 1 columns of Q. No, the first n - 1 columns of Q will be computed using normal Gram-Schmidt, but another column cannot be constructed. Yes, the first n - 1 columns of Q will be computed using normal Gram-Schmidt, then we can choose a unit vector orthogonal to the previous columns to construct the last column of Q. No, the method will fail completely, a QR factorization cannot be computed.

Explanation / Answer

Ans-

Consider the GramSchmidt procedure, with the vectors to be considered in the process as
columns of the matrix A. That is,
Then,
A =
·
u
u
u
1
2
k+1
a
1
¯
¯
= a
= a
= a
a
2
1
2
¯
¯
¢ ¢ ¢
¯
; e
¡ (a
k+1
Note that jj ¢ jj is the L
2
1
¡ (a
1.1 QR Factorization
2
¯
a
n
¸
:
=
u
¢ e
1
jju
)e
k+1
norm.
1
1
1
jj
;
; e
¢ e
1

)e
1

2
=
u
¡ ¢ ¢ ¢ ¡ (a
jju
2
2
jj
:
k+1
¢ e
k

)e
k

; e
k+1
=
u
jju
k+1
k+1
jj
:

The resulting QR factorization is
A =
·
a
1
¯
¯
a
2
¯
¯
¢ ¢ ¢
¯
Note that once we ¯nd e
¯
a
n
¸
=
1
·
e
; : : : ; e
1
n
¯
¯
e
2
¯
¯
¢ ¢ ¢
¯
¯
e
n

¸
2
6
6
6
4
a

1

¢ e
1

0 a
.
.
.
a
2
2
¢ e
¢ e
.
.
.
1
2
¢ ¢ ¢ a
¢ ¢ ¢ a
.
.
.
0 0 ¢ ¢ ¢ a
, it is not hard to write the QR factorization.
n
n
n
¢ e
¢ e
.
.
.
¢ e
1
2
n
3
7
7
7
5
= QR:

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